## Linear Approximation (Linearization) and Differentials

Linear approximation, sometimes called linearization, is one of the more useful applications of tangent line equations. We can use linear approximations to estimate the value of more complex functions. Coming up with a linear function that closely approximates another function at a certain point gives you something that is a lot easier to work with than the original function.

## How to find a linear approximation at a point

To find the linear approximation of a function at a point, you can simply use the formula for the linearization of a function. Let’s say you are given a function $$f(x)=x^3+2x$$ and you want to find its liner approximation at the point $\mathbf{a=-2}$. You can do this by applying the formula

$$L(x)=f(a)+f'(a)(x-a).$$

The first step is to find the derivative of f(x). In this case we can find f'(x) by using the power rule.

$$f'(x)=3x^2+2$$

Now that you know f(x), f'(x), and the value of a, you can plug all of this into the linearization formula above to find the linear approximation of f(x) near x=a.

$$L(x)=f(a)+f'(a)(x-a)$$ $$L(x)=\big( (-2)^3+2(-2) \big)+ \big( 3(-2)^2+2 \big) \big( x-(-2) \big)$$ $$L(x)=\big( -12 \big)+ \big( 14 \big) \big( x+2 \big)$$ $$L(x)=-12 + 14x+28$$ $$L(x)=14x+16$$

## Useful Applications of Linear Approximation

At this point you’re probably thinking, “this is just finding tangent line equations, who cares?”

But this does have some useful applications. One such application is using linear approximation to estimate the value of numbers that would be difficult or impossible to find without a calculator.

Let me show you what I mean by that.

### Example

Find the linearization of $\mathbf{f(x)=\sqrt{x+1}}$ at a=0 and use it to approximate $\mathbf{f(x)=\sqrt{0.9}}$.

Finding $\mathbf{f(x)=\sqrt{0.9}}$ would be difficult without a calculator, right? But with linear approximation we can get pretty close by applying the process above and taking it a step further pretty easily.

We already have been given f(x) and the a value we need, so you just need to find f'(x) to apply our linearization formula.

$$f(x)=\sqrt{x+1}$$ $$f(x)=(x+1)^{\frac{1}{2}}$$

Writing f(x) in this form, we can simply apply the power rule and chain rule to find f'(x).

$$f'(x)=\frac{1}{2}(x+1)^{-\frac{1}{2}}$$ $$f'(x)=\frac{1}{2\sqrt{x+1}}$$

Now that we know f'(x), let’s first use that and f(x) to calculate f(a) and f'(a) before applying all of this to the linearization formula. Remember we were given a=0.

$$f(x) = \sqrt{x+1}$$ $$f(0)=\sqrt{0+1}$$ $$f(0)=\sqrt{1}$$ $$f(0)=1$$

And now you can find f'(a), or f'(0).

$$f'(x)=\frac{1}{2\sqrt{x+1}}$$ $$f'(0)=\frac{1}{2\sqrt{0+1}}$$ $$f'(0)=\frac{1}{2\sqrt{1}}$$ $$f'(0)=\frac{1}{2}$$

Now we can plug all of this into our linearization formula.

$$L(x)=f(0)+f'(0)(x-0)$$ $$L(x)=1+\frac{1}{2}x$$

#### How do you use this to estimate a number?

So we know that the function $L(x)=1+\frac{1}{2}x$ is a good estimate of $f(x)=\sqrt{x+1}$ when we plug in numbers close to a=0 for x. And if we plug x=0 into f(x), that would give us $f(0)=\sqrt{0+1}=\sqrt{1}$ which is close to the number we are trying to estimate, $f(x)=\sqrt{0.9}$.

In reality, we could find the exact value of $\sqrt{0.9}$ by evaluating $f(-0.1)=\sqrt{-0.1+1}=\sqrt{0.9}$. But this is difficult, so instead we can use our linear approximation because we know it’s close to f(x) for x values near x=0. So instead of finding f(-0.1) we can find L(-0.1) and that will give us a good estimate.

$$L(-0.1)=1+\frac{1}{2}(-0.1)$$ $$L(-0.1)=1+(0.5)(-0.1)$$ $$L(-0.1)=1-0.05$$ $$L(-0.1)=0.95$$

So this tells us that $\sqrt{0.9} \approx 0.95$.

## What if you aren’t given a function f(x) or an x=a value to use?

This seems nice to be able to estimate complicated values using this technique, but what if you need to estimate a complex value and aren’t given a function to use. Sometimes you aren’t given a function. And if you were trying to use this technique in the real world you probably wouldn’t be given a nice function to use and an a value that’s close to the input you need to estimate.

So how do you deal with this?

Well, the short answer is that you need to come up with the function and the a value on your own. Then once you have these you can apply the same technique above. Let me show you what I mean.

Let’s say you are told:

Use linear approximation to estimate $\sqrt{24}$.

The first thing you want to do is come up with the function to use to apply the linearization formula to. Since we are trying to find $\sqrt{24}$, our function is clearly going to need a square root in it somewhere. It is possible that you may need to experiment a bit and see how it works out, so let’s just start with the most obvious choice for now and say $f(x)=\sqrt{x}$.

If we do say $f(x)=\sqrt{x}$ then we can see that $f(24)=\sqrt{24}$ which is exactly the number that we are trying to estimate. But f(24) is hard to evaluate. So to find the a value we will want to use, you should consider what number is near 24 that we could plug into the function $f(x)=\sqrt{x}$ and easily figure out the result?

The closest number to 24 that we can easily plug into our function would be 25. Because 25 is a perfect square, we know that $f(25)=\sqrt{25}=5$, which is a nice round number. This is perfect for linear approximation because we need an input that is near the location we are trying to estimate whose output we know.

Since 25 is near 24 and we know the exact value of f(25), we can find the linear approximation of $f(x)=\sqrt{x}$ at a=25 and use this to estimate $f(24)=\sqrt{24}$ just like we did in the example above.

I’m not going to work this all the way through because it will be the exact same process as the last example now that we know $f(x)=\sqrt{x}$ and a=25. You can also see the rest of the solution to this problem in the video above.

## Differentials and how to find dy

Differentials go hand-in-hand with linear approximation and they have some interesting applications of their own.

Given some function f(x), you can find its differential, dy, simply by using the following formula.

$$dy=f'(x) \ dx$$

For example, say you are given the function $y=e^{\frac{x}{10}}$ and you are told to find the differential dy. All you need to do is apply the above formula.

You know that $$f(x)=e^{\frac{x}{10}}$$ Therefore, you can apply the chain rule to find $$f'(x)=e^{\frac{x}{10}} \cdot \frac{1}{10}$$ $$f'(x)=\frac{1}{10}e^{\frac{x}{10}}$$

So as a result, we know that

$$dy=\frac{1}{10}e^{\frac{x}{10}} \ dx$$

## Application of Differentials: Estimating Error

Estimating error is one of the more common uses for differentials. This can then be used to find the percentage error of a measurement. Let me explain what this means with an example.

Let’s say you measure a sphere to have a radius of 5 m. However, you know that the instrument you used to measure the radius of this sphere could have up to 5 cm (or 0.05 m) of error. Use differentials to estimate the maximum error in the measured volume of this sphere.

Knowing the the volume of a sphere is $$V=\frac{4}{3}\pi r^3$$ we will use this as our function. We will be able to use the differential of this function to estimate the maximum possible error in the measured volume of this sphere based on the maximum possible error in the radius. We already know the possible error in the radius, so we can use this and relate it to the volume.

First all we need to do is find the differential of our volume equation. This will require taking the derivative of $\mathbf{V=\frac{4}{3}\pi r^3}$. Using the formula of a differential as stated in the previous section, the differential is

$$dV=4 \pi r^2 \ dr$$

Then with this, all you need to do is plug in the information we know to find the possible error in the volume. Keep in mind, r represents the radius of the sphere that we know to be 5 m. And dr will represent the possible error in the radius, which we know to be 0.05 m. It’s important that we use the same units for all of these inputs since the output will use the same units in that case.

$$dV=4 \pi (5m)^2(0.05m)$$ $$dV=5 \pi m^3$$

So we know that the maximum error in the measurement we have for the volume of this sphere is $\mathbf{5 \pi m^3\approx 15.708m^3}$.

## Percentage Error

Once we find the maximum possible error as outlined above, we can use this to find the percentage error of our measurement. This is simply the maximum possible error in the measurement divided by the measurement itself. So in this case it would be

$$Percentage \ Error = \frac{error \ in \ volume}{volume}.$$

$$Percentage \ Error = \frac{5 \pi}{\frac{500}{3}\pi} = 3\%$$

## Implicit Differentiation Examples

If you haven’t already read about implicit differentiation, you can read more about it here. Once you check that out, we’ll get into a few more examples below.

$\mathbf{1. \ \ ycos(x) = x^2 + y^2}$ | Solution

$\mathbf{2. \ \ xy=x-y}$ | Solution

$\mathbf{3. \ \ x^2-4xy+y^2=4}$ | Solution

$\mathbf{4. \ \ \sqrt{x+y}=x^4+y^4}$ | Solution

$\mathbf{5. \ \ e^{x^2y}=x+y}$ | Solution

For each of the above equations, we want to find dy/dx by implicit differentiation. In general a problem like this is going to follow the same general outline. Although, this outline won’t apply to every problem where you need to find dy/dx, this is the most common, and generally a good place to start. Start with these steps, and if they don’t get you any closer to finding dy/dx, you can try something else. Here are the steps:

• Take the derivative of both sides of the equation with respect to x.
• Separate all of the dy/dx terms from the non-dy/dx terms.
• Factor out the dy/dx.
• Isolate dy/dx.

Some of these examples will be using product rule and chain rule to find dy/dx.

## Solutions

$\mathbf{1. \ \ ycos(x) = x^2 + y^2}$

$$ycos(x)=x^2+y^2$$ $$\frac{d}{dx} \big[ ycos(x) \big] = \frac{d}{dx} \big[ x^2 + y^2 \big]$$ $$\frac{dy}{dx}cos(x) + y \big( -sin(x) \big) = 2x + 2y \frac{dy}{dx}$$ $$\frac{dy}{dx}cos(x) – y sin(x) = 2x + 2y \frac{dy}{dx}$$ $$\frac{dy}{dx}cos(x) -2y \frac{dy}{dx} = 2x + ysin(x)$$ $$\frac{dy}{dx} \big[ cos(x) -2y \big] = 2x + ysin(x)$$ $$\frac{dy}{dx} = \frac{2x + ysin(x)}{cos(x) -2y}$$

$\mathbf{2. \ \ xy=x-y}$

$$xy = x-y$$ $$\frac{d}{dx} \big[ xy \big] = \frac{d}{dx} \big[ x-y \big]$$ $$1 \cdot y + x \frac{dy}{dx} = 1-\frac{dy}{dx}$$ $$y+x \frac{dy}{dx} = 1 – \frac{dy}{dx}$$ $$x \frac{dy}{dx} + \frac{dy}{dx} = 1-y$$ $$\frac{dy}{dx} \big[ x+1 \big] = 1-y$$ $$\frac{dy}{dx} = \frac{1-y}{x+1}$$

$\mathbf{3. \ \ x^2-4xy+y^2=4}$

$$x^2-4xy+y^2=4$$ $$\frac{d}{dx} \big[ x^2-4xy+y^2 \big] = \frac{d}{dx} \big[ 4 \big]$$ $$2x \ – \bigg[ 4x \frac{dy}{dx} + 4y \bigg] + 2y \frac{dy}{dx} = 0$$ $$2x \ – 4x \frac{dy}{dx} – 4y + 2y \frac{dy}{dx} = 0$$ $$-4x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x+4y$$ $$\frac{dy}{dx} \big[ -4x+2y \big] = -2x+4y$$ $$\frac{dy}{dx}=\frac{-2x+4y}{-4x+2y}$$ $$\frac{dy}{dx}=\frac{-x+2y}{-2x+y}$$

$\mathbf{4. \ \ \sqrt{x+y}=x^4+y^4}$

$$\sqrt{x+y}=x^4+y^4$$ $$\big( x+y \big)^{\frac{1}{2}}=x^4+y^4$$ $$\frac{d}{dx} \bigg[ \big( x+y \big)^{\frac{1}{2}}\bigg] = \frac{d}{dx}\bigg[x^4+y^4 \bigg]$$ $$\frac{1}{2} \big( x+y \big) ^{-\frac{1}{2}} \bigg( 1+\frac{dy}{dx} \bigg)=4x^3+4y^3\frac{dy}{dx}$$ $$\frac{1}{2} \cdot \frac{1}{\sqrt{x+y}} \cdot \frac{1+\frac{dy}{dx}}{1} = 4x^3+4y^3\frac{dy}{dx}$$ $$\frac{1+\frac{dy}{dx}}{2 \sqrt{x+y}}= 4x^3+4y^3\frac{dy}{dx}$$ $$1+\frac{dy}{dx}= \bigg[ 4x^3+4y^3\frac{dy}{dx} \bigg] \cdot 2 \sqrt{x+y}$$ $$1+\frac{dy}{dx}= 8x^3 \sqrt{x+y} + 8y^3 \frac{dy}{dx} \sqrt{x+y}$$ $$\frac{dy}{dx} \ – \ 8y^3 \frac{dy}{dx} \sqrt{x+y}= 8x^3 \sqrt{x+y} \ – \ 1$$ $$\frac{dy}{dx} \bigg[ 1 \ – \ 8y^3 \sqrt{x+y} \bigg]= 8x^3 \sqrt{x+y} \ – \ 1$$ $$\frac{dy}{dx}= \frac{8x^3 \sqrt{x+y} \ – \ 1}{1 \ – \ 8y^3 \sqrt{x+y}}$$

$\mathbf{5. \ \ e^{x^2y}=x+y}$

$$e^{x^2y}=x+y$$ $$\frac{d}{dx} \Big[ e^{x^2y} \Big] = \frac{d}{dx} \big[ x+y \big]$$ $$e^{x^2y} \bigg( 2xy + x^2 \frac{dy}{dx} \bigg) = 1 + \frac{dy}{dx}$$ $$2xye^{x^2y} + x^2e^{x^2y} \frac{dy}{dx} = 1+ \frac{dy}{dx}$$ $$x^2e^{x^2y} \frac{dy}{dx} \ – \ \frac{dy}{dx} = 1 \ – \ 2xye^{x^2y}$$ $$\frac{dy}{dx} \big(x^2e^{x^2y} \ – \ 1 \big) = 1 \ – \ 2xye^{x^2y}$$ $$\frac{dy}{dx} = \frac{1 \ – \ 2xye^{x^2y}}{x^2e^{x^2y} \ – \ 1}$$

The top of a ladder slides down a vertical wall at a rate of 0.15 m/s. At the moment when the bottom of the ladder is 3 m from the wall, it slides away from the wall at a rate of 0.2 m/s. How long is the ladder?

This is a fairly common example of a related rates problem and a common application of derivatives and implicit differentiation. I’m sure you may have come across a related rates ladder problem like this. If I can offer one piece of advice for this type of problem it’d be this: don’t use this ladder, it always falls…

Alright, bad jokes aside, this is going to follow the same 4 steps as all the other related rates problems I’ve done. If you’d rather watch a video, then check out my video below. But otherwise, let’s jump into it with the usual process!

## 1. Draw a sketch

As always, we’ll start by drawing a quick sketch of all of the information that is being described in the problem. To do this we should first think about what information we have. First of all, we need to think about the shape that’s being formed with the ladder.

Since the ladder is standing on the ground and leaning up against a vertical wall, we can say that a triangle would be formed by the 3 objects in the problem. More specifically we know that the vertical wall forms a 90 degree angle with the ground. Therefore, the triangle formed by the ground, the wall, and the ladder would be a right triangle.

On top of this, the problem also gives us a few pieces of information about the dimensions of the triangle and how they are changing. It actually tells us about how fast the ladder is moving, but since the ladder is what forms the triangle, we can deduce how the dimensions of the triangle are changing.

We are given 3 pieces of information about the position of the ladder as well as how the ladder is moving at the specific instant we are looking at.

• Bottom of the ladder is 3 m away from the wall.
• Top of the ladder is moving down the wall at a rate of 0.15 m/s.
• Bottom of the ladder is moving away from the wall at a rate of 0.2 m/s.

Adding these labels to our drawing from above would give us something like this:

This sketch gives us a pretty good idea of what is going on in this problem. Not only that, but we will be able to use this to get an idea of what kind of equation we will need to come up with.

## 2. Come up with your equation

Before we come up with our equation we want to sort through the information we are given and asked to find. This is important because we need to decide what measurements and variables we want in the equation.

#### What are we looking for?

The question asks us to find the length of the ladder. Therefore, we will need to find the length of the hypotenuse of the triangle in our drawing. Because of this we want to be sure to include the hypotenuse of our triangle in our equation.

#### What do we know about?

Looking back up at our labeled drawing, you can see that we really only have information about the bottom and side of our triangle. We know the length of the bottom side of the triangle and the rate of change of this side.

And we were also given information about the rate of change of the left side of the triangle. But remember that our equation in this step cannot include rates of change. Instead, the fact that we know this rate of change tells us that we can use the left side length of the triangle in our equation, not its rate of change.

We aren’t given any information about the angles in the triangle other than the fact that it’s a right triangle. As a result, we probably don’t want our equation to involve the angles of this triangle.

Since we know that our equation either needs to include, or can include the lengths of the sides of the triangle we should label them. Let’s go back to our drawing and label the side of the triangle. You can label them whatever you’d like, but I’ll go with x, y, and z.

#### Putting it into an equation

At this point we’ve figured out that we need an equation that related the sides of a right triangle.

What do you know about the relationship between the sides of a right triangle, but neither of the other two angles?

You’re probably thinking Pythagorean Theorem. If you are, you’re right! Pythagorean Theorem tells us that the square of the hypotenuse is equal to the sum of the squares of the other two sides of a right triangle. Remember this can only be applied to the sides of a right triangle, so noticing that is actually very important. In other words we know $$z^2=x^2+y^2.$$

## 3. Implicit differentiation

Now that we have come up with our equation, we need to apply implicit differentiation to take the derivative of both sides of our equation with respect to time.

$$\frac{d}{dt} \Big[ z^2 = x^2 + y^2 \Big]$$

Before we do this though I want to point something out. Let’s look at each of the letters in this equation and consider how we need to treat them when we differentiate with respect to time.

Consider z first. z represents the length of the hypotenuse of the triangle. This is the side that is formed by the ladder. As time changes what happens to the length of the ladder? Nothing. It doesn’t change at all. It’s constant. Therefore we can treat z like a constant. If z is a constant and never changes, then $z^2$ would be constant too. It doesn’t change as time changes.

So when we take the derivative of z with respect to time, the derivative will be 0. The derivative of any constant is 0.

Unfortunately x and y won’t be as convenient. Looking back at our drawings you can see that the sides labelled x and y are changing over time. As the ladder slides down and away from the wall, these two sides of the triangle change in length. Therefore, when we take the derivative of $x^2$ and $y^2$ we will need to treat x and y as functions of time. Doing this means that we will need to use the chain rule, where x and y are the inside function and they are being plugged into another function that squares them.

#### Back to the derivative

Knowing how to treat each letter in our equation, let’s go ahead and take the derivative of both sides with respect to time.

$$\frac{d}{dt} \Big[ z^2 = x^2 + y^2 \Big]$$ $$0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$

## 4. Solve for the desired rate of change

Now all we need to do is plug in all of the information we have and solve for the right variable. However, this one is a little weird. The reason I say this is that we are actually not looking for a rate of change.

Remember the question told us to find the length of the ladder. Which means we need to find the value of z. The differential equation we just ended up with doesn’t even have a z in it, so how can we use it to find z?

Well, we’re actually going to need to go back to our original equation. $$z^2=x^2+y^2$$ We know the value of x based on information we were given, but we don’t know the value of y yet. If we could figure out what y was, then we could use this equation, plug in the value for x and y, then solve for z.

#### How do we find y?

This is what we will use our differential equation from the previous step for. That equation has a y in it, and we know the value of all the other variables.

• We are told that the moment we are considering is when the bottom of the ladder is 3 m from the wall. Since that corresponds to the side of our triangle labelled x, we know $\mathbf{x=3}$.
• We are also told that the ladder is moving away from the wall at a rate of 0.2 m/s. Therefore, x must be getting longer, or increasing, at that rate. So $\mathbf{\frac{dx}{dt}=0.2}$.
• And finally, the ladder is sliding down the wall at a rate of 0.15 m/s. So y must be getting shorter, or decreasing, at that rate. This means $\mathbf{\frac{dy}{dt}=-0.15}$.

#### Plugging it all into our equation

Knowing all of the values in our equation aside from y, we can plug these in and solve for y.

$$0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$ $$0 = 2(3)(0.2) + 2y(-0.15)$$ $$0=1.2-0.3y$$ $$0.3y=1.2$$ $$y=4$$

Now that we know x and y, we can plug them back into our original equation and solve for z.

$$z^2=x^2+y^2$$ $$z^2=(3)^2+(4)^2$$ $$z^2=25$$ $$z=5$$

So the ladder must be 5 m long!

## L’HOSPITAL’S RULE – HOW TO – With Examples

L’Hospital’s Rule really just tells us one thing that makes evaluating certain limits a lot easier. Limits that meet 3 specific requirements can be made much simpler using L’Hospital’s Rule. First let me introduce L’Hospital’s Rule, then we can go over the 3 conditions that you need to check before you can apply it to any given limit.

## What does L’Hospital’s Rule tell us?

To find a limit of a function that is a fraction, we can take the derivative of the top of the fraction and the derivative of the bottom of the fraction and make a new fraction out of the derivatives.

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

Notice that we don’t use the quotient rule here. The reason for this is that we are taking the limit of this fraction and not taking the derivative of this fraction. The limit is the key piece that allows us to avoid the quotient rule and take the derivative of each piece of the fraction separately to create another limit that is equal to the original.

The hope is that the fraction resulting from the derivatives will be easier to evaluate than the original limit was.

## How do we know when to use L’Hospital’s Rule?

Before we can apply L’Hospital’s rule to any given limit, we need to confirm that these three conditions are met:

1. f(x) and g(x) are differentiable on some open interval that includes $\mathbf{x=a}$. This will basically just mean that both the numerator and denominator are differentiable at $x=a$.
2. $\mathbf{g'(x) \neq 0}$ near $\mathbf{x=a}$. Note that it doesn’t matter if $g'(x)=0$ AT $x=a$ as long as you can pick some interval (as small as is necessary) around $x=a$ where $g'(x) \neq 0$ for all x‘s in that interval besides $x=a$. You likely won’t need to worry about running into a function that you can’t pick a small enough interval around $x=a$ to make this work.
3. As x $\rightarrow$ a, f(x) AND g(x) $\mathbf{\rightarrow 0}$ — OR — f(x) AND g(x) $\mathbf{\rightarrow \pm \infty}$

That’s really all there is to it. Let’s jump into some practice problems and I will show you how to apply L’Hospitals Rule.

## Example 1

Evaluate the following limit using L’Hospital’s Rule or explain why L’Hospital’s Rule cannot be used to evaluate this limit.

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x}$$

If we tried to use limit properties to evaluate this limit, we would see that both the top and the bottom of this fraction go to either positive or negative infinity as x goes to infinity.

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x} \rightarrow \frac{\infty}{- \infty}$$

#### So L’Hospital’s Rule might help…

Now we just need to confirm that the other two conditions are met.

$f(x)=e^x$ is an exponential function and is differentiable everywhere, for any value of x. And $g(x)=-x^2 + 1000x$ is also differentiable everywhere since it’s a polynomial. Since it’s differentiable everywhere, it is also differentiable for any infinitely large x value.

$g'(x) = -2x+1000$ will go to $-\infty$ as x approaches $\infty$. $g'(x) \neq 0$ for any infinitely large x value since it just continues to go to $- \infty$.

So we know that this limit meets all 3 requirements needed to apply L’Hospital’s Rule.

#### Now we know we can apply L’Hospital’s Rule

Taking the derivative of the top and bottom of the fraction individually tells us that:

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x} = \lim_{x \to \infty} \frac{e^x}{-2x+1000}$$

Now we can evaluate this new limit instead. But if we do this, we will notice that we will still end up in the same situation that we had before.

$$\lim_{x \to \infty} \frac{e^x}{-2x+1000} \rightarrow \frac{\infty}{- \infty}$$

#### But what if it didn’t really help make the limit easier?

Which puts us in a perfect situation to consider using L’Hospital’s Rule to evaluate this new limit as well. By the same logic as before, we can confirm that the first two conditions are met as well as x gets infinitely large. Since all 3 required conditions are met we can go ahead and apply L’Hospital’s Rule a second time.

$$\lim_{x \to \infty} \frac{e^x}{-2x+1000} = \lim_{x \to \infty} \frac{e^x}{-2}$$

Now we can simply use the basic limit properties to evaluate this last limit.

#### Now we have a much easier limit

$$\lim_{x \to \infty} \frac{e^x}{-2} = \ – \frac{1}{2} \lim_{x \to \infty} e^x = \ – \infty$$

So therefore,

$$\lim_{x \to \infty} \frac{e^x}{-x^2+1000x} = \ – \infty$$

#### A quick note on applying L’Hospital’s Rule twice

This is an interesting problem because it shows that you can apply L’Hospital’s Rule multiple times on the same problem. You just need to make sure that each time you apply it, the resulting limit still meets all 3 required conditions before applying it to the new limit. There is no limit to the number of times you can continue applying L’Hospital’s Rule over and over in the same problem as long as you are making sure that the limit you are applying it to meets all 3 conditions every time you apply it.

## Example 2

Evaluate the following limit using L’Hospital’s Rule or explain why L’Hospital’s Rule cannot be used to evaluate this limit.

$$\lim_{x \to 3} \frac{x-3}{27-x^3}$$

Again, if we think about what value the top and bottom of this fraction will go towards as x approaches 3, we would see that

$$\lim_{x \to 3} \frac{x-3}{27-x^3} \rightarrow \frac{0}{0}$$

Since we get another indeterminate form, which is $\frac{0}{0}$, we should consider using L’Hospital’s Rule to make this limit easier to evaluate.

#### So L’Hospital’s Rule might help…

First, we need to make sure that the other two conditions are met as well.

We can check that $g'(x) = -3x^2$ doesn’t equal zero anywhere near $x=3$. This is because $g'(x) = -3x^2$ is continuous everywhere and the only place where $g'(x)=-3x^2=0$ is when $x=0$. As a result of these two things, we can pick some interval around $x=3$ that doesn’t include $x=0$ to satisfy condition #2.

Also, both f(x) and g(x) are polynomials and are therefore differentiable everywhere. So we know they will both be differentiable on any interval around $x=3$.

#### Now we know we can apply L’Hospital’s Rule

Doing so by taking the derivative of the top and bottom of our fraction separately tells us that

$$\lim_{x \to 3} \frac{x-3}{27-x^3} = \lim_{x \to 3} \frac{1}{-3x^2}$$

And doing this gives us an easier limit to deal with. Now we can simply apply the limit properties to evaluate. Applying the limit properties tells us that:

$$\lim_{x \to 3} \frac{1}{-3x^2} = \frac{1}{-3 \Big( \lim_{x \to 3}x \Big) ^2}= \frac{1}{-3(3)^2} = \ – \frac{1}{27}$$

So therefore we know that:

$$\lim_{x \to 3} \frac{x-3}{27-x^3} = \ – \frac{1}{27}$$

## Example 3

Evaluate the following limit using L’Hospital’s Rule or explain why L’Hospital’s Rule cannot be used to evaluate this limit.

$$\lim_{x \to 0} \frac{|x|}{x^5+2x}$$

If we think about what value the numerator and denominator of this fraction will approach as x approaches 0 from both sides, we would see:

$$\lim_{x \to 0} \frac{|x|}{x^5+2x} \rightarrow \frac{0}{0}$$

Since we get an indeterminate form, which is $\frac{0}{0}$, we should consider using L’Hospital’s Rule to make this limit easier to evaluate.

#### So L’Hospital’s Rule might help…

First, we need to make sure that the other two conditions are met as well.

Upon checking condition #1 however, we run into a problem. Condition #1 requires that f(x) and g(x) are both differentiable on some interval containing $x=0$, including $x=0$.

But $f(x) = |x|$ is not differentiable at $x=0$. Therefore, we actually can’t apply L’Hospital’s Rule to evaluate this limit. I won’t go into the details here since we won’t be using L’Hospital’s Rule. But if you want to try evaluating this limit, I’d recommend considering both one-sided limits on their own and compare them to start. You can see a similar application here.

## Example 4

$\mathbf{\lim\limits_{x \to \infty} \frac{ln(x)}{\sqrt{x}}}$ | Solution

## Optimization Problems Part 2

In my last lesson, I introduced optimization problems and I discussed local extrema. You should check that out if you haven’t already. The next thing that I would like to discuss now is finding global maximums and minimums. The first step in finding a global maximum or minimum of a function is actually very similar to finding the local max and min values.

#### But once you know about the local maximums and minimums, how do you find the global extrema?

Finding the global extrema from the local extrema is really quite simple. And there really is only one way to find the global maximum and minimum values of a function. You just need to find a list of all possible x values where the global max or global min may occur. Then once you have created a list of all possibilities, you just plug them all into the original function.

Not the function’s derivative or the function’s second derivative. But the original function.

You will test for global extrema of f(x) using f(x), not f'(x) or f”(x). This is after you have your list of all possible locations where the global extrema could occur (which will require the use of f'(x)). But my point is that there wouldn’t be a first derivative test or a second derivative test with the global extrema like there was with finding local extrema.

## So what does this look like in practice?

Let’s use an example. For example, let’s say that we are asked to find the global maximum and the global minimum of $f(x)=2x^3-\frac{5}{2}x^2+x-1$ on the domain $-4 \leq x \leq 5$.

Notice we are being asked to find the global extrema on a specific domain. This is important because a lot of functions either go to infinity or to negative infinity as x either gets infinitely large, infinitely small, or approaches some specific value. So if we were asked to find the global maximum of a function that goes to infinity as x goes to infinity, we wouldn’t be able to do this. There would be no maximum since the function only continues to grow.

So we know we will be limited to a specific domain.

As I said before, finding global extrema starts out exactly like finding local extrema. The first thing we need to do is find the critical values of our function. To do this, we just need to find its derivative and set $f'(x)=0$ and solve for x.

$$f(x)=2x^3-\frac{5}{2}x^2+x-1$$ $$f'(x)=6x^2-5x+1$$ Then set $$6x^2-5x+1=0$$ and solve for x to find the critical values. To do this, we can factor the left side of the equation. $$(3x-1)(2x-1)=0$$ To solve this we can set each factor equal to zero individually. $$3x-1=0 \ \ \ \ and \ \ \ \ 2x-1=0$$ $$3x=1 \ \ \ \ and \ \ \ \ 2x=1$$ $$x=\frac{1}{3}, \ \frac{1}{2}$$

So now we know that this function has two critical values, $x= \frac{1}{3}$ and $x=\frac{1}{2}$. Now this is where things get different with a global max/min problem versus finding the local max/min. We also need to consider the endpoints of our given domain as critical values!

This will always be the case when we are looking for a global maximum or minimum. The problem asked us to find the global extrema on the domain $-4 \leq x \leq 5$. Therefore, we will also say that $x=-4$ and $x=5$ will be treated as critical values that we need to test.

### So how to we test our critical values?

The first thing that I would like to do is list out all of the x values we will be testing in one place. Remember, the list of values we need to test came from two places:

1. Setting $f'(x)=0$ and solving for x.
2. Each of the endpoints of the domain on which we need to find the global maximums and minimums.

So in this case we’ll have four total x values that we need to test: $$x=-4, \ \frac{1}{3}, \ \frac{1}{2}, \ and \ 5$$ To test these points, all we need to do is plug each of the four points into f(x). Whichever on outputs the largest number will tell us the global maximum. Whichever outputs the smallest number will tell us the global minimum.

$$f(-4)= \ 2(-4)^3-\frac{5}{2}(-4)^2+(-4)-1 \ = -173$$ $$f \bigg( \frac{1}{3} \bigg) = \ 2 \bigg( \frac{1}{3} \bigg) ^3-\frac{5}{2} \bigg( \frac{1}{3} \bigg) ^2+ \bigg( \frac{1}{3} \bigg) -1 \ = -\frac{47}{54}$$ $$f \bigg( \frac{1}{2} \bigg) = \ 2 \bigg( \frac{1}{2} \bigg) ^3-\frac{5}{2} \bigg( \frac{1}{2} \bigg) ^2+ \bigg( \frac{1}{2} \bigg) -1 \ = -\frac{7}{8}$$ $$f(5)= \ 2(5)^3-\frac{5}{2}(5)^2+(5)-1 \ = \frac{383}{2}$$

So we can see that the smallest of these four numbers is -173 and the largest of them is $\frac{383}{2}$. Therefore, the global maximum of f(x) on $-4 \leq x \leq 5$ is $\frac{383}{2}$ which occurs when $x=5$. And the global minimum of f(x) on $-4 \leq x \leq 5$ is -173 which occurs when $x=-4$.

We can even graph our function using Desmos along with the critical points to make sure our answer makes sense. You can click on the link in the last sentence to see a larger version of the graph.

## Extra practice

If you’d like some extra practice finding global maximums and minimums, here are a few examples you can work through on your own. A couple of these examples will require the use of the product rule and quotient rule.

For each of the following, find the global maximum and minimum of the given function on the given domain or explain why one doesn’t exist.

$$f(x)= 2x^4 + 5x^2 – 12x \ \ \textrm{on the domain} -1 \leq x \leq 2$$ $$g(x)= xe^x +6x^3-12 \ \ \textrm{on the domain} -3 \leq x \leq 0$$ $$h(x)= \frac{x^4-3x^2+1}{x+1} \ \ \textrm{on the domain} -2 \leq x \leq \frac{3}{2}$$ $$j(x)= \frac{x^4-3x^2+1}{x+1} \ \ \textrm{on the domain} -\frac{9}{10} \leq x \leq \frac{3}{2}$$

Hopefully all of this helps with global maximums and minimums, but as always I’d love to hear your questions if you have any. If you find that you get stuck as you’re working through some of these extra practice problems just let me know. Just email me at jakesmathlessons@gmail.com and I’ll see if I can help provide a bit more clarification. You can also use the form below to subscribe to my email list and I will send you my bonus FREE calculus 1 study guide to help you survive calculus! Just put in your name and email address and I’ll be sure to let you know when I post new content! Feel free to go check out my other lessons and solutions about derivatives as well.

## Optimization Problems

Optimization problems are another common application of the derivative. Usually in these problems you are given some function or described some situation and are then asked to find different maximums and minimums. There are a few different things that are commonly asked that you optimize, so I’d like to go over the different categories with you.

## Local maximums and minimums

The most common thing that comes up in optimization problems is finding the local maximums and minimums. The best way to do this is using the derivative of the function you are trying to optimize. Taking the function’s derivative will tell you everything you need to know.

In order to find the list of all x values where the local extrema may occur, you just need to take the function’s derivative, set it equal to zero, and solve for x. In other words, you can find the x values that will give you local max and min values by setting up the equation $$f'(x)=0$$ and solving it for x. Keep in mind, this equation often has multiple solutions, so make sure you include all possible solutions.

Doing this will give you a list of x values where all possible local maximums and minimums occur. These are called critical numbers.

Once you have your list of critical points, you will often need to figure out which ones are maximums and minimums. There are two tests you can conduct to figure which one they are.

### First derivative test

This is usually the method I like to use. As you might guess, the first derivative test only requires the use of the first derivative. I usually use this test because we already had to find the first derivative to get our list of critical values.

The first thing I would suggest doing before beginning your test is drawing out a number line and putting your critical values on it. Let’s just say for example that we had some function f(x), took its derivative, and found that the critical values are $x=-1, \ 2, \ 6$. Our number line might look something like this.

That’s all you need on your number line at this point. Don’t label any extra x values besides your critical values. It will only make things more confusing later.

Now all we need to do is plug x values into f’ that are around these critical values to figure out where f is increasing and decreasing. So we will need to plug in some number that is in each of the following 4 intervals. $$x<-1,$$ $$-1 < x < 2,$$ $$2 < x < 6,$$ $$x>6.$$ So all we need to do is just plug in some number in each segment of our labeled number line.

It doesn’t matter which number you plug in from each of those intervals, so you can pick whichever numbers seem easiest to plug into f’. Let’s say we will plug $x=-2, \ 0, \ 4, \ 7$ into f’. We want to plug them into f’ because we are trying to figure out information about the slope of f. This will tell us where it’s increasing and decreasing. Let’s imagine we plug these four x values into f’ and find that $$f'(-2)=-4, \ \ \ f'(0)=6, \ \ \ f'(4)=2, \ \ \ f'(7)=-7.$$

We only really care if these values are positive or negative. If f’ is positive at a certain x value, we know f must have a positive slope. And if f’ is negative at a certain x value, we know f must have a negative slope.

Since f'(-2) is negative, f must have a negative slope at $x=-2$. And f must also have a negative slope for all $x<-1$ since that is the interval from our number line that $x=-2$ falls within. So we should label this interval with a negative slope, like this:

Then we want to do the same thing for the interval of $-1 \leq x \leq 2$. We found out that $f'(0)=6$, which is a positive number. Therefore, f must have a positive slope for all $-1 \leq x \leq 2$. So we can label our number line accordingly.

Then we want to do the same thing with the other two intervals. This would give us something like this for our number line:

Now we just need to use this number line to determine which critical values are maximums and which are minimums. There are really only 3 main cases you need to think about for each critical value.

1. If f is increasing to the left and decreasing to the right, that critical point will be a local maximum. This will cause this little section of the graph to look like a frowny face. Therefore, the critical point will higher than the graph right around it.
2. If f is decreasing to the left and increasing to the right, that critical point will be a local minimum. This will cause this little section of the graph to look like a smiley face. Therefore, the critical point will lower than the graph right around it.
3. If f is decreasing to the left and the right or if it’s increasing to the left and the right, that critical point will NOT be a local maximum or a local minimum.

So let’s look at each of our three critical values on the number line above and see which category they all fall into.

• For $x=-1$, you can see that f is decreasing on the left side and increasing on the right side. Therefore, the section of the f(x) right around $x=-1$ looks like a smiley face and would be a local minimum.
• For $x=2$, you can see that f is increasing on the left side and increasing on the right side. Therefore, the critical value $x=2$ would not be a local maximum or a local minimum. We would need f(x) to change direction at $x=2$ for it to be a maximum or minimum, but that doesn’t happen here.
• For $x=6$, you can see that f is increasing on the left side and decreasing on the right side. Therefore, the section of the f(x) right around $x=6$ looks like a frowny face and would be a local maximum.

### Second derivative test

The other way you can test to see if each critical value is a local maximum or a local minimum is with the second derivative test. You do not need to use both methods if you are only trying to find local extrema because they will give you the same conclusions. Just pick which test you like more. This method will require us to find the second derivative of our function, or f”(x). We can find this simply by finding the derivative of f'(x), which we already found.

Just like with the first derivative test, it helps to draw everything out on a number line. Start with just drawing a number line that only contains the critical values which we found a while ago.

Now we need to plug each of our critical values into our second derivative, or f”(x). One important difference is that we had to plug numbers around our critical values with the first derivative test. But with the second derivative test, we will actually plug in the critical values instead of numbers around them.

Since we need to plug each critical value into our second derivative, this means we will plug $x=-1, \ \ 2, \ \ 6$ into f”(x). When we do that, let’s imagine we find that $$f”(-1) = 2, \ \ f”(2) = 0, \ \ f”(6) = -9.$$

Just like before, it doesn’t really matter what the exact values are that we just found. All that matters is whether they are positive, negative, or zero. If f” is positive at a certain point, then f would be concave up at that point. And if f” is negative at a point, then f is concave down at that point. If f” is zero, then f isn’t concave up or concave down at that point.

Since f”(-1) is positive (we just found that it’s 2), we know that f is concave up when $x=-1$. That just means that it’s curved upward, like a smiley face. So we can indicate this on our number line to keep track of what we have so far.

Since f”(2) is zero, we know that f is not concave up or down when $x=2$. This tells us that $x=2$ is the point where f switches from being concave up to concave down, or vise versa. Since f doesn’t have concavity (curvature) here, we will show this as a flat line on our number line.

And lastly, since f”(6) is negative (we just found that it’s -9), we know that f is concave down when $x=6$. That just means that it’s curved downward, like a frowny face. Therefore, we might get something like this.

So now we just need to figure out what all this means when it comes to the second derivative test. Again, there are three cases we want to look for.

1. If f(x) is concave up, or f”(x) is positive, for some critical value x, then this critical value represents a local minimum.
2. If f(x) is concave down, or f”(x) is negative, for some critical value x, then this critical value represents a local maximum.
3. If f(x) isn’t concave up or down, or f”(x) is zero, for some critical value x, then this critical value could be a local minimum or local maximum or neither.

So let’s compare these to our critical values to see if they are each local maximums or minimums.

• For $x=-1$, you can see that f is concave up. Therefore, the section of the f(x) right around $x=-1$ looks like a smiley face and would be a local minimum.
• For $x=2$, you can see that f is not concave up or concave down. In this case we don’t know from the second derivative test if this critical value would be a local maximum or a local minimum.
• For $x=6$, you can see that f is concave down. Therefore, the section of the f(x) right around $x=6$ looks like a frowny face and would be a local maximum.

Notice that these are the exact same results we found from the first derivative test, aside from the undetermined critical value. I know we didn’t actually have a function for f(x) to work through, but you would find the same thing if you did actually go through these processes with some function. To find which critical values are local maximums, local minimums, or neither, you only need to do one of these two tests.

### Extra practice

Find the critical values for the following functions and determine whether each one is a local minimum, local maximum, or neither. A couple of these examples will require the use of the product rule and the quotient rule, so check those out if you need a refresher.

$$f(x)= 2x^4 + 5x^2 – 12x$$ $$g(x)= xe^x +6x^3-12$$ $$h(x)= \frac{x^4-3x^2+1}{x+1}$$

Hopefully all of this helps you gain a bit of a better understanding of local extrema, but as always I’d love to hear your questions if you have any. Go check out part 2 of my coverage on optimization problems where I go over global maximums and minimums.

## How to find the equation of a tangent line

One common application of the derivative is to find the equation of a tangent line to a function. Usually when you’re doing a problem like this, you will be given a function whose tangent line you need to find. And you will also be given a point or an x value where the line needs to be tangent to the given function.

Using these two pieces of information, you need to create an equation for a line that satisfies the required conditions. This process is very closely related to linear approximation (or linearization) and differentials.

When coming up with the equation of the line, there are a couple different approached you could take. You should decide which one to use based on your own personal preference. The only difference between the different approaches is which template for an equation of a line you prefer to use. Remember, there are two main forms that a line will take: $$y=mx+b$$ $$y=m(x-x_0)+y_0$$ Another thing to keep in mind is that the first form is generally easier when we are given the y-intercept of the line. The second form above is usually easier when we are given any other point that isn’t the y-intercept.

In both of these forms, x and y are variables and m is the slope of the line. In the first equation, b is the y-intercept. And in the second equation, $x_0$ and $y_0$ are the x and y coordinates of some point that lies on the line. This could be any point that lies on the line.

It is also important to notice that a line would be tangent to a function at a specific point if and only if the following two conditions are met.

• The function and its tangent line need to go through the same point
• and they both need to share the same slope at that shared point.

## But how does the derivative apply?

You should always keep in mind that a derivative tells you about the slope of a function. So if we take a function’s derivative, then look at it at a certain point, we have some information about the slope of the function at that point. Since a tangent line has to have the same slope as the function it’s tangent to at the specific point, we will use the derivative to find m.

So let’s jump into a couple examples and I’ll show you how to do something like this.

## Example 1

Find the equation of the tangent line to the function $\mathbf{y=x^3+4x-6}$ at the point (2, 10).

In order to find this tangent line, let’s consider the two conditions that need to be met for our line to be a tangent line at the specified point.

1. The tangent line and the given function need to go through the same point. Since the problem told us to find the tangent line at the point $(2, \ 10)$, we know this will be the point that our line has to go through.
2. The tangent line and the function need to have the same slope at the point $(2, \ 10)$. In order to find this slope we will need to use the derivative. Let’s start with this.

#### Finding the slope of the tangent line

Remember that the derivative of a function tells you about its slope. So to find the slope of the given function $y=x^3+4x-6$ we will need to take its derivative. This will just require the use of the power rule. $$y’=3x^2+4$$

But how can we use this to find the slope of the tangent line when it has variables in it?

This is where the specific point we need to consider comes into play. We know that the tangent line and the function need to have the same slope at the point $(2, \ 10)$. Therefore, they need to have the same slope when $x=2$. In order to find the slope of the given function y at $x=2$, all we need to do is plug 2 into the derivative of y.

Therefore, the slope of our line would simply be $$y'(2)=3(2)^2+4=16.$$ And because of this we also know the slope of our tangent line will be $$m=16.$$ So we know this will guarantee that our tangent line has the right slope, now we just need to make sure it goes through the right point.

#### Making sure the tangent line contains the given point

Since we do know a point that has to lie on our line, but don’t know the y-intercept of the line, it would be easier to use the following form for our tangent line equation. $$y=m(x-x_0)+y_0$$

And since we already know $m=16$, let’s go ahead and plug that into our equation. $$y=16(x-x_0)+y_0$$

Now to finish our tangent line equation, we just need the x and y coordinates of a point that lies on this line. Then we can simply plug them in for $x_0$ and $y_0$. Well, we were given this information! We were told that the line we come up with needs to be tangent at the point $(2, \ 10)$. Therefore, our tangent line needs to go through that point. This tells us our tangent line equation must be $$y=16(x-2)+10$$ $$y=16x-32+10$$ $$y=16x-22$$

And that’s it! We know that the line $y=16x-22$ will go through the point $(2, 10)$ on our original function. And we know that it will also have the same slope as the function at that point.

We can even use Desmos to check this and see what our function and tangent line look like together.

## Example 2

Find the equation of the line that is tangent to the function $f(x) = xe^x$ when $x=0$.

To start a problem like this I suggest thinking about the two conditions we need to meet.

1. The tangent line and the given function need to intersect at $\mathbf{x=0}$. This time we weren’t given the y coordinate of this point so we will need to figure that out.
2. Then we need to make sure that our tangent line has the same slope as f(x) when $\mathbf{x=0}$.

#### Finding the slope of the tangent line

I personally think that it’s a little easier to find the slope of the tangent line first, but you can start with making sure the other condition is met if you prefer.

When you’re asked to find something to do with slope, your first thought should be to use the derivative. The derivative of a function tells you about it’s slope. Since we need the slope of f(x), we’ll need its derivative.

Looking at our function $f(x)=xe^x$ you can see that it is the product of two simpler functions. To find it’s derivative we will need to use the product rule. I’m not going to show every step of this, but if you aren’t 100% sure how to find this derivative you should click the link in the last sentence. $$f'(x) = e^x + xe^x$$ $$f'(x) = e^x \big(1+x \big)$$

Now consider the fact that we need our tangent line to have the same slope as f(x) when $x=0$. To find the slope of f(x) at $x=0$ we just need to plug in 0 for x into the equation we found for f'(x). $$f'(0) = e^{(0)} \big( 1 + (0) \big)$$ $$f'(0) = 1(1)=1$$

So we know that the slope of our tangent line needs to be 1.

#### Making sure the tangent line contains the required point

Now we just need to make sure that our tangent line shares the same point as the function when $x=0$. In order to do this, we need to find the y value of the function when $x=0$. This would be the same as finding f(0). $$f(0) = (0)e^{(0)} = 0$$

Since this is the y value when $x=0$, we can also say that this is the y-intercept. We know the y intercept of our tangent line is 0. Since we figured out the y-intercept, it would be easiest to use the $y=mx+b$ form of the line for the tangent line equation.

We already found that the slope will be 1 and that the y-intercept will need to be 0, so we can plug these values in for m and b. Doing this tells us that the equation of our tangent line is $$y=(1)x+(0)$$ $$y=x.$$

Again, we can see what this looks like and check our work by graphing these two functions with Desmos.

## Finding the Tangent Line Equation with Implicit Differentiation

Depending on the curve whose tangent line equation you are looking for, you may need to apply implicit differentiation to find the slope.

## Example 3

Find the equation of the line that is tangent to the curve $\mathbf{y^3+xy-x^2=9}$ at the point (1, 2).

First we need to apply implicit differentiation to find the slope of our tangent line.

$$\frac{d}{dx} \big[ y^3 + xy – x^2 \big] = \frac{d}{dx} [9]$$ $$3y^2 \frac{dy}{dx} + 1\cdot y + x \cdot \frac{dy}{dx} – 2x = 0$$ $$3y^2 \frac{dy}{dx} + x \frac{dy}{dx} = -y + 2x$$ $$\frac{dy}{dx} \big[ 3y^2 + x \big] = -y + 2x$$ $$\frac{dy}{dx} = \frac{-y+2x}{3y^2+x}$$

Now we can plug in the given point (1, 2) into our equation for $\mathbf{\frac{dy}{dx}}$ to find the slope of the tangent line.

$$m=\frac{-(2)+2(1)}{3(2)^2+(1)}=\frac{0}{13}=0$$

With this slope, we can go back to the point slope form of a line. Since we know the slope and a point that lies on this line, we can plug that information into the general point slope form for a line. This will leave us with the equation for a tangent line at the given point.

$$y=m(x-x_0)+y_0$$ $$y=0(x-1)+2$$ $$y=2$$

So the constant function $\mathbf{y=2}$ is tangent to the curve $\mathbf{y^3+xy-x^2=9}$ at the point (1, 2).

## Example 4

Find the equation of the line that is tangent to the curve $\mathbf{16x^2 + y^2 = xy + 4}$ at the point (0, 2).

Again, we will start by applying implicit differentiation to find the slope of the tangent line.

$$\frac{d}{dx} \big[ 16x^2 + y^2 \big] = \frac{d}{dx} [xy + 4]$$ $$32x + 2y \frac{dy}{dx} = 1\cdot y + x \cdot \frac{dy}{dx}$$ $$2y \frac{dy}{dx} – x \frac{dy}{dx}= -32x + y$$ $$\frac{dy}{dx} \big[ 2y-x \big] = -32x+y$$ $$\frac{dy}{dx} = \frac{-32x+y}{2y-x}$$

Now we can plug in the given point (0, 2) into our equation for $\mathbf{\frac{dy}{dx}}$ to find the slope of the tangent line.

$$m = \frac{-32(0)+(2)}{2(2)-(0)}$$ $$m=\frac{2}{4}$$ $$m=\frac{1}{2}$$

With this slope, we can go back to the point slope form of a line. Since we know the slope and a point that lies on this line, we can plug that information into the general point slope form for a line. This will leave us with the equation for a tangent line at the given point.

$$y=m(x-x_0)+y_0$$ $$y=\frac{1}{2}(x-0)+2$$ $$y=\frac{1}{2}x+2$$

## Tangent Line Equation Without Derivatives

There are some cases where you can find the slope of a tangent line without having to take a derivative. This is not super common because it does require being able to take advantage of additional information. Usually you will be able to do this if you know some geometrical fact about the curve whose tangent line equation you are looking for.

The most common example of this is finding the a line that is tangent to a circle.

## Example 5

Find the equation of the line that is tangent to the circle $\mathbf{(x-2)^2+(y+1)^2=25}$ at the point (5, 3).

We already are given a point that we know needs to lie on our tangent line. This tells us that if we can find the slope of the tangent line, we would just be able to plug it all into the point slope form for a linear function and we would have a tangent line. So we just need to find the slope of the tangent line.

In order to find this slope we can take advantage of a geometrical fact about circles: a line connecting the center of a circle to its edge will be perpendicular to a line that is tangent to the circle at that edge point. Below you can see what this looks like on a graph of this circle, or at least a portion of it.

Therefore, if we know the slope of a line connecting the center of our circle to the point (5, 3) we can use this to find the slope of our tangent line. Based on the general form of a circle, we know that $\mathbf{(x-2)^2+(y+1)^2=25}$ is the equation for a circle that is centered at (2, -1) and has a radius of 5. So we need to find the slope of a line connecting the points (5, 3) and (2, -1). We can do this using the formula for the slope of a line between two points.

$$slope = \frac{y_2 – y_1}{x_2 – x_1}$$ $$slope = \frac{3 – (-1)}{5 – 2}$$ $$slope = \frac{4}{3}$$

Now we can simply take the negative reciprocal of $\mathbf{\frac{4}{3}}$ to find the slope of our tangent line. So we know the slope of our tangent line will be $\mathbf{- \frac{3}{4}}$.

Knowing that the slope of our tangent line will be $\mathbf{- \frac{3}{4}}$ and that it will go through the point (5, 3), we can put this into the point slope form of a line to find the equation of our tangent line.

$$y=m(x-x_0)+y_0$$ $$y= \ – \frac{3}{4}(x-5)+3$$ $$y= \ – \frac{3}{4}x + \frac{15}{4}+3$$ $$y= \ – \frac{3}{4}x + \frac{15}{4}+ \frac{12}{4}$$ $$y= \ – \frac{3}{4}x + \frac{27}{4}$$

Hopefully all of this helps you gain a bit of a better understanding of finding tangent lines, but as always I’d love to hear your questions if you have any. Just email me at jakesmathlessons@gmail.com and I’ll see if I can help provide a bit more clarification. You can also use the form below to subscribe to my email list and I’ll send you my FREE bonus study guide to help you survive calculus! Just put in your name and email address and I’ll be sure to let you know when I post new content! Feel free to go check out my other lessons and solutions about derivatives as well.

## Examples of product, quotient, and chain rules

I have already discuss the product rule, quotient rule, and chain rule in previous lessons. But I wanted to show you some more complex examples that involve these rules. The reason for this is that there are times when you’ll need to use more than one of these rules in one problem. So let’s dive right into it!

## Example 1

Find the derivative of $y \ = \ sin(x^2 \cdot ln \ x)$.

At first glance of this problem, the first thing we should notice is that we can think of this function as one function plugged into another. There is a clear inner function and a clear outer function.

It can be broken down as $x^2 \cdot ln \ x$ being plugged into sin(x) for x. Since our function can be thought of as one function plugged into another, we will want to start out with the chain rule.

### Chain rule

The first thing we need to do to apply the chain rule is to figure out our inside function and outside function. It’s usually easier to think about the insider function first.

#### Finding f and g

To find the inside function we just need to answer the question: what function is being plugged into another?

Looking at our function you can see that we are taking $x^2 \cdot ln \ x$ and plugging that into another function. So we will say $$g_1(x) = x^2 \cdot ln \ x$$

Now that we have decided on the inside function, we need to find the outside function. All we need to do here is look at the original function, and replace our inside function with a single x. So we will replace $x^2 \cdot ln \ x$ with just x. This gives us $$f_1(x) = sin(x).$$

#### Finding f’ and g’

Now that we have found f and g, we just need to take each of their derivatives to find f’ and g’.

Finding f’ should be simple here. $$f’_1(x) = cos(x)$$

Finding g’ will be a little more tricky.

Looking at our function $g_1(x)$ you can see that it is actually the product of two simpler functions, $x^2$ and $ln \ x$. Therefore, we are going to have to use the product rule to find this derivative. You can kind of think of this as a smaller sub-problem within our problem, so we will come back to the chain rule after applying the product rule.

### Product rule

We need to use the product rule to find the derivative of $$g_1(x) = x^2 \cdot ln \ x.$$ The product rule starts out similarly to the chain rule, finding f and g. However, this time I will use $f_2(x)$ and $g_2(x)$.

#### Finding f and g

With the product rule it doesn’t really matter which function is f and which is g. As long as we correctly identify that our function is a product of two simpler functions, it’ll work out correctly. So we will say $$f_2(x) = x^2$$ $$g_2(x) = ln \ x.$$

#### Finding f’ and g’

Now we just need to find the derivatives of f and g. Since they are fairly simple functions, this shouldn’t be too difficult.

To find the derivative of f we just need to use the power rule. $$f’_2(x) = 2x.$$

And finding the derivative of g should be a derivative that you have memorized. Using Wolfram Alpha we can see that $$g’_2(x) = \frac{1}{x}.$$

#### Plugging into the formula

Now that we have found all the pieces we need, we can simply plug them all into the product rule formula. $$g’_1(x) = f_2(x) \cdot g’_2(x) \ + \ f’_2(x) \cdot g_2(x)$$ $$\frac{d}{dx} \Big[ x^2 \cdot ln \ x \Big] \ = \ x^2 \cdot \frac{1}{x} \ + \ 2x \cdot ln \ x$$ $$\frac{d}{dx} \Big[ x^2 \cdot ln \ x \Big] \ = \ x \ + \ 2x \cdot ln \ x$$

### Back to chain rule

Now that we know the derivative of $x^2 \cdot ln \ x$ we can go back up to the chain rule. We knew that $$g_1(x) = \ x^2 \cdot ln \ x$$ and by using the product we just found that $$g’_1(x) = \ x \ + \ 2x \cdot ln \ x.$$ Just as a quick reminder, we already found that $$f_1(x) = sin(x)$$ $$f’_1(x) = cos(x).$$ Now we just need to plug these four pieces into the formula for chain rule. $$h’_1(x) = \ f’_1 \big( g_1(x) \big) \cdot g’_1(x)$$ $$\frac{d}{dx} \Big[ sin \big(x^2 \cdot ln \ x \big) \Big] \ = \ cos \big( x^2 \cdot ln \ x \big) \cdot \big( x \ + \ 2x \cdot ln \ x \big)$$

And that’s it! We could factor out a like term out of one of these factors, but it wouldn’t really make the function any simpler so I won’t do that. We can also use Wolfram Alpha to check our answer. A quick note on that, Wolfram Alpha uses “log” instead of “ln” to describe a natural log.

## Example 2

Find the derivative of $y = \frac{x \ sin(x)}{ln \ x}$.

Looking at this function we can clearly see that we have a fraction. Therefore, we can break this function down into two simpler functions that are part of a quotient. So we can see that we will need to use quotient rule to find this derivative.

### Quotient rule

As discussed in my quotient rule lesson, when we apply the quotient rule to find a function’s derivative we need to first determine which parts of our function will be called f and g.

#### Finding f and g

With the quotient rule, it’s fairly straight forward to determine which part of our function will be f and which part will be g. We will always say f is the numerator (top of our fraction) and g is the denominator (bottom of our fraction). So we can say $$f_1(x) \ = \ x \ sin(x)$$ $$g_1(x) \ = ln \ x.$$

#### Finding f’ and g’

Once we have determined which part of our function we are going to call f and which part will be g, we need to take each of their derivatives so we can use the quotient rule formula.

First we’ll start with finding f’. To find this we need to find the derivative of $f_1(x)= x \ sin(x)$. Notice this function is actually a product of two simpler functions. So in order to find $\mathbf{f'_1(x)}$ we will actually need to use the product rule. This will create a smaller sub-problem for us so we will need to come back to the quotient rule in a moment.

### Product rule

As we did in the previous example, or in my product rule lesson, we need to start by determining which piece of the function $f_1(x) = x \ sin(x)$ will be $f_2$ and which will be $g_2$.

#### Finding f and g

When using the product rule it doesn’t really matter which piece of the product is called f and g. So we will say $$f_2(x) = x$$ $$g_2(x) = sin(x).$$

#### Finding f’ and g’

In order to use the product rule formula, we need to find the derivative of each of these pieces now. Fortunately, both of these pieces are simple functions to differentiate. $$f’_2(x) = 1$$ $$g’_2(x) = cos(x)$$

#### Plugging into the formula

Now we just need to plug the four pieces we’ve found into the product rule formula. $$f’_1(x) = \ f_2(x) \cdot g’_2(x) \ + \ f’_2(x) \cdot g_2(x)$$ $$\frac{d}{dx} \Big[ x \ sin(x) \Big] \ = \ x \cdot cos(x) \ + \ 1 \cdot sin(x)$$ $$\frac{d}{dx} \Big[ x \ sin(x) \Big] \ = \ x \ cos(x) \ + \ sin(x)$$

### Back to the quotient rule

Now that we have used the product rule to find $$f’_1(x) = \ x \ cos(x) \ + \ sin(x)$$ we need to find $g'_1(x)$ so we can use the quotient rule formula. Remember $g_1(x) = ln \ x$, which is a function whose derivative you should memorize. $$g’_1(x) = \frac{1}{x}.$$ So now we know all of the pieces we need to apply the quotient rule formula.

$$h'(x) \ = \ \frac{f’_1(x) \cdot g_1(x) \ – \ f_1(x) \cdot g’_1(x)}{g^2_1(x)}$$ $$\frac{d}{dx} \Bigg[ \frac{x \ sin(x)}{ln \ x} \Bigg] \ = \ \frac{ \Big[ \big( x \ cos(x) + sin(x) \big) \cdot ln \ x \Big] \ – \ \Big[ x \ sin(x) \cdot \frac{1}{x} \Big]}{\big( ln \ x \big)^2}$$

And now we can just simplify by distributing through all of our parenthesis.

$$\frac{d}{dx} \Bigg[ \frac{x \ sin(x)}{ln \ x} \Bigg] \ = \ \frac{ x \ ln(x) \ cos(x) \ + \ ln(x) \ sin(x) \ – \ sin(x)}{ln^2(x)}$$

And that’s the answer! Again, we can check this using Wolfram Alpha.

If you have any questions on any of this just let me know! You can email me at jakesmathlessons@gmail.com. You can also use the contact form below and I’ll add you to my email list and send you my calculus 1 study guide to help you boost your calculus scores! I’d also love to hear any suggestions for future posts so please don’t hesitate to reach out to me. If you want some more practice with derivatives go check out my other lessons and problems related to derivatives.

## RELATED RATES – Triangle Problem (changing angle)

A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground. When the angle of elevation is $\pi$/3, this angle is decreasing at a rate of $\pi$/6 rad/min. How fast is the plane traveling at this time?

## 1. Draw a sketch

We are going to go ahead and proceed with the 4 steps that I use for all related rates problems. You can check those out in my related rates lesson. As with any related rates problem, the first thing we should do is draw a sketch of the situation being described in this problem.

In this drawing the plane is moving from left to right. Keep in mind that the problem also said that the angle $\theta$ is $\frac{\pi}{3}$ radians and is decreasing at a rate of $\frac{\pi}{6} \ \frac{rad}{min}$. I did not label these facts, but they are important and we should remember them.

## 2. Come up with your equation

Now that we have made our sketch, we need to use this to come up with an equation relating the necessary quantities and measurements. When doing this we need to consider what the question is asking us to find and what information was given.

However, this may be difficult with the drawing we have so far. First we will need to create and label some other measurements based on some geometric shape we already know something about. Looking back at the initial sketch, it seems like our situation can be described with a triangle. So let’s consider the following drawing instead.

Notice that I have added a few things that weren’t in the first drawing. First of all, it is much more clear that we can look at this situation as a triangle. Also I have added labels to the bottom side and hypotenuse of this triangle.

Keep in mind that the side labeled as 5 km will measure the height of the plane as it moves to the right. Therefore, it will always maintain a right angle with the ground. It will also always be 5 km because the altitude of the plane is not changing as it flies horizontally.

Now let’s consider how to make an equation out of this triangle.

### What are we looking for?

The question that this problem asks us is “how fast is the plane traveling at this time?” Another way to think about how fast the plane is traveling is how quickly its position is changing, or the rate of change of the plane’s position.

In order to measure the plane’s position at different times, we need to measure how far away it is from some stationary point. This is exactly what the tracking telescope is (shown as the black semi-circle on the ground in our drawing). Since the plane is not moving exactly in the opposite direction of the tracking telescope, we can’t simply use the side of the triangle that measures the distance between these two things (the side labeled z).

However, since the plane is moving horizontally, there is another distance we have labeled that can be used. Since it’s moving horizontally, we will need to use a horizontal distance. In our drawing this is exactly what side x represents! It is the horizontal distance between the telescope and the point directly beneath the plane!

So the rate of change of the plane’s position is represented by the rate of change of the horizontal distance between the plane and the telescope. Therefore, the rate of change of x will give us our answer. Since we will introduce the rate of change when we take the derivative of our equation, we just need to be sure to include x in our initial equation.

### What do we know about?

The problem gives us a few pieces of information that we may need to know.

• Plane’s altitude is always 5 km (it won’t change because it says the plane flies horizontally)
• Angle of elevation is $\frac{\pi}{3}$ radians
• Angle of elevation is decreasing at a rate of $\frac{\pi}{6} \ \frac{rad}{min}$

In general, the angle of elevation is by definition the angle between the object being measured and the horizontal. In this case we made the ground perfectly horizontal to make this easier to visualize, but as long as we measure against some horizontal line the ground doesn’t have to be flat.

So to summarize, we know that the right side of the triangle will be a constant 5 km. We also have some information about our angle $\mathbf{\theta}$ and about its rate of change.

I also want to point out what we can figure out if needed. Since we know two of the three angles, we could use this to find the third. Knowing all three angles and the length of a side, we can use these to find either of the other two sides. This means we can find x or z at this instant if we need to, so they are fine to use in our equation.

### Putting it into an equation.

Up to this point we have sorted through what we need to find, which told us that we need to use x in our equation. We also know that we can use $\theta$, z, or the constant 5 km side. So let’s think about what we can do with this.

You may be thinking that we should use Pythagorean Theorem at this point, so let’s think about this. If we do use Pythagorean Theorem our equation will use all three side lengths and will not use our angle $\theta$. We don’t necessarily have to use our angle, so there’s nothing wrong with this.

However, think about what will happen when we take the derivative of this equation with respect to time. At this point we would introduce $\frac{dz}{dt}$ and $\frac{dx}{dt}$! Since we don’t know either of these values, this would leave us with two unknowns in only one equation, which isn’t very helpful because it doesn’t allow us to solve for the one variable we want.

#### But what else can we do?

Well the last thing we tried didn’t include our angle $\theta$, so let’s think about what we can do with that. What function do you know about that deals with an angle of a right triangle and any number of its sides?

Sine, cosine, and tangent!

Remember soh, cah, toa? We need to use one of these here. Keep in mind that each of these uses one angle and two of the sides of the triangle. Since we already know some information about $\mathbf{\theta}$ and its rate of change, we will use that as our angle. But how do you decide which sides to use?

We already know that we need to use x. We decided that several paragraphs ago (click here to go back up there). So this leaves the side z and the constant 5 km side to chose from for our other side. If we chose z, we will run into the same problem as if we use Pythagorean Theorem, we don’t know anything about it’s rate of change!

Notice since the 5 km side is constant, we don’t need to worry about the rate of change. This side is a constant so its rate of change would be 0! So we will use x and the constant 5 km side in our equation. Now there’s just one last thing to figure out.

#### Do we use sine, cosine, or tangent?

When deciding between sine, cosine, and tangent we will want to use soh, cah, toa:

Sine: Opposite / Hypotenuse

We know that we are using the angle $\theta$, and the x side and the constant 5 km side. Let’s think for a second about where these sides lie in relation to the angle we’re looking at.

First of all, neither of these two sides are the hypotenuse. The hypotenuse of a right triangle is the longest side which is opposite from the right angle. Therefore, one will be the side adjacent to $\theta$ and the other will be opposite to $\theta$. In general, the adjacent side would be the one that’s touching $\theta$. In this case, the adjacent side would be x. The opposite side would be the one side that is not touching our angle, making the 5 km side opposite.

So we need to chose between sine, cosine, and tangent so that we incorporate the adjacent side and the opposite side. Using soh, cah, toa, we can see that tangent uses the opposite and adjacent sides. This tells us that taking the tangent of our angle gives us the opposite side divided by the adjacent side. $$tan(\theta) = \frac{\textrm{opposite}}{\textrm{adjacent}}$$ $$tan(\theta) = \frac{5}{x}$$

## 3. Implicit differentiation

Now that we have come up with our equation, we need to take its derivative with respect to time. This will require the use of the chain rule since we are differentiating with respect to time. The reason for this is that we need to treat $\theta$ and x as functions of time.

Notice that our equation has a fraction. As a result we would need to use the quotient rule to find the derivative of the right side of our equation. However, we can rewrite this in another form to make the derivative a bit simpler.

Remember, we can always move a term from the bottom of a fraction up to the top by making its power negative. Therefore, we can write our equation as $$tan(\theta) = 5x^{-1}.$$

Now that we just have a constant multiplied by an x term to a constant power, we can use the power rule instead of the quotient rule on the right side of the equation. We do still have to use the chain rule, but at least we made things a bit simpler. In order to find the derivative of $tan(\theta)$ we would need to use the fact that $tan(x)=\frac{sin(x)}{cos(x)}$ then use the quotient rule. I’m not going to show this process, but instead I’ll use Wolfram Alpha to find that $$\frac{d}{dx} tan(x) = sec^2(x) = \frac{1}{cos^2(x)}.$$

Applying this to our problem we can implicitly differentiate our equation. $$\frac{d}{dt} \big[ tan(\theta) \big]= \frac{d}{dt} \Big[ 5x^{-1} \Big]$$ $$sec^2(\theta) \cdot \frac{d \theta}{dt} = -5x^{-2} \cdot \frac{dx}{dt}$$

## 4. Solve for desired rate of change

Remember the thing we need to find in this problem is the rate of change of x. So all we need to do now is isolate $\frac{dx}{dt}$ because that’s exactly what represents the rate of change of x. $$sec^2(\theta) \cdot \frac{d \theta}{dt} = -5x^{-2} \cdot \frac{dx}{dt}$$ $$-\frac{1}{5} sec^2(\theta) \cdot \frac{d \theta}{dt} = x^{-2} \cdot \frac{dx}{dt}$$ $$-\frac{1}{5} x^2 \cdot sec^2(\theta) \cdot \frac{d \theta}{dt} = \frac{dx}{dt}$$

And once we’ve done that, we just need to plug in all of the other variables to find $\frac{dx}{dt}$! The other variables that we’ll need to plug in are x, $\theta$, and $\frac{d \theta}{dt}$.

We already know from the actual problem that “the angle of elevation is $\pi$/3,” and “this angle is decreasing at a rate of $\pi$/6 rad/min.” This directly tells us that $$\theta = \frac{\pi}{3}.$$ And since this angle is decreasing we know its rate of change will be negative, so $$\frac{d \theta}{dt} = -\frac{\pi}{6}.$$

Now we just need to find x. Since it wasn’t directly given, finding x will require a little work but it’s not too complicated.

We will actually need to use the equation we created before taking its derivative. $$tan(\theta) = \frac{5}{x}$$ Since we need to find x at the moment when $\theta = \frac{\pi}{3}$ we can plug this in here to find x. $$tan \bigg( \frac{\pi}{3} \bigg) = \frac{5}{x}$$ $$\frac{sin(\pi/3)}{cos(\pi/3)} = \frac{5}{x}$$

And now using the unit circle. $$\frac{\sqrt{3}/2}{1/2} = \frac{5}{x}$$ $$\sqrt{3} = \frac{5}{x}$$ $$\sqrt{3} \cdot x = 5$$ $$x = \frac{5}{\sqrt{3}}$$

#### Plugging them all in

$$\frac{dx}{dt} = -\frac{1}{5} x^2 \cdot sec^2(\theta) \cdot \frac{d \theta}{dt}$$ $$\frac{dx}{dt} = -\frac{1}{5} \bigg( \frac{5}{\sqrt{3}} \bigg)^2 \cdot sec^2\bigg( \frac{\pi}{3} \bigg) \cdot \bigg( -\frac{\pi}{6} \bigg)$$ $$\frac{dx}{dt} = -\frac{1}{5} \bigg( \frac{25}{3} \bigg) \cdot \frac{1}{cos^2\big( \frac{\pi}{3} \big)} \cdot \bigg( -\frac{\pi}{6} \bigg)$$ $$\frac{dx}{dt} = -\frac{1}{5} \bigg( \frac{25}{3} \bigg) \cdot \frac{1}{1/4} \cdot \bigg( -\frac{\pi}{6} \bigg)$$ $$\frac{dx}{dt} = -\frac{1}{5} \bigg( \frac{25}{3} \bigg) \cdot 4 \cdot \bigg( -\frac{\pi}{6} \bigg)$$ $$\frac{dx}{dt} = \frac{100\pi}{90} = \frac{10\pi}{9}$$

So we know that the plane is traveling at a speed of $\mathbf{\frac{10\pi}{9} \ \frac{km}{min}}$!

Hopefully that all helped this problem make a little more sense. If you still want some more practice with triangle related rates problems, check some of these out:

Solution – At noon, ship A is 150 km west of ship B.  Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h.  How fast is the distance between the ships changing at 4:00 PM?

Solution – A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?

Or you can check out my related rates lesson where I have a list of other related rates problems with solutions if you want more practice with these.

If you have any questions about this problem, please let me know! I’d be happy to help. You can email any questions to me at jakesmathlessons@gmail.com or use the form below to join my email list and I’ll send you my calc 1 study guide as a welcome gift!

## RELATED RATES – Square Problem

Each side of a square is increasing at a rate of 6 $\frac{cm}{s}$. At what rate is the area of the square increasing when the area of the square is 16 $cm^2$?

If you haven’t already, you should check out my related rates lesson. I go through the steps that can be used to solve any related rates problem. I will use these steps here, but it might be useful for you to see a more detailed explanation of why we use these steps.

## 1. Draw a sketch

The first thing we will always want to do is draw a sketch of the situation described by the problem. This problem is relatively simple to draw out. Here we have a square whose sides are growing at a constant rate. We know the sides are growing at a rate of 6 $\frac{cm}{s}$ and we want to consider the moment when the area of the square is 16 $cm^2$.

## 2. Come up with your equation

Now that we have our drawing laid out, we need to create an equation whose derivative we will need to find.

#### What are we looking for?

This question is asking us to find the rate at which the area of the square is increasing. So it tells us that we need to have some variable that represents the rate of change of the area at some point.

Of course, we will need to take the derivative of our equation soon. Doing this will introduce the ‘rate of change’ part. Therefore, we need to make sure that our equation we make includes the area of the square. As long as we do this, we will end up with the rate of change of the area once we take its derivative.

#### What do we know about?

There was only two pieces of information that the question directly told us.

• Each side of a square is increasing at a rate of 6 $\frac{cm}{s}$.
• The area of the square in this instant is 16 $cm^2$.

#### Putting it into an equation

So at this point, we have figured out that we need our equation to include the square’s area, and we know something about the square’s area and the rate of change of its sides.

Although, we don’t know anything about the actual side lengths at the given instant, we can figure that out if we have to. Therefore, it’s fine if our equation includes the length of the square’s sides.

To summarize, the only measurements of this square we have discussed so far are its side lengths and its area. So our equation should relate its area and its side lengths. Keep in mind, we don’t want to put anything that represents a rate of change into our equation. These will come when we take the derivative of our equation.

The simplest equation that relates the area of a square with its side lengths would likely be the formula for the area of a square that you are already familiar with. $$A=l^2$$ Where A is the area of the square, and l is the length of its sides.

## 3. Implicit differentiation

Once we have created our equation like we did above, we need to go ahead and take its derivative with respect to time. This will be done using implicit differentiation.

Since we will be taking the derivative with respect to time, we will need to treat the A and the l in our equation as functions of time. This will require using the chain rule to find their derivatives.

$$\frac{d}{dt} \big[ A \big] = \frac{d}{dt} \Big[ l^2 \Big]$$ $$\frac{dA}{dt} = 2l \cdot \frac{dl}{dt}$$

## 4. Solve for the desired rate of change

Now all we have to do is solve for the rate of change the question is asking about. Just like I said earlier, we need to find the rate of change of the square’s area. Since this is exactly what $\frac{dA}{dt}$ represents and we have already isolated this, we don’t have much else to do. All we need to do is plug in the values we have for everything else in our equation.

The only other variables we need to plug in are l and $\frac{dl}{dt}$. The tricky thing here is that we don’t directly know what l is at the moment in this problem. Instead we know that the area of the square is
16 $cm^2$. Since we know that the relationship between the area of a square and its side lengths is $$A=l^2$$ we can find l in this instant. $$16=l^2$$ $$4=l$$ So we know that in this moment the length of the square’s sides is 4 cm.

Remember, the problem also told us that the side lengths are increasing at a rate of 6 $\frac{cm}{s}$. This directly tells us the rate of change of the sides lengths. So we also know that $$\frac{dl}{dt}=6$$

Putting all of this into our equation will give us: $$\frac{dA}{dt} = 2l \cdot \frac{dl}{dt}$$ $$\frac{dA}{dt} = 2(4)(6)$$ $$\frac{dA}{dt} = 48$$

So the area of this square is increasing at a rate of 48 $\mathbf{\frac{cm^2}{s}}$ when the area is 16 $\mathbf{cm^2}$.

If you’re still having some trouble with related rates problems or just want some more practice you should check out my related rates lesson. At the bottom of this lesson there is a list of related rates problems that I have posted along with their solutions. I also have several other lessons and problems on the derivatives page you can check out. If you can’t find the topic or question you’re looking for just let me know by emailing me at jakesmathlessons@gmail.com!

You can also enter your name and email using the form below and I will send you my calculus 1 study guide as a welcome gift!