## Implicit Differentiation Practice Problem

Find $$\frac{dy}{dx}$$ if $$y=x^x$$.

## Solution

This is kind of a tricky problem. Obviously, if we need to find $$\frac{dy}{dx}$$, we need to take the derivative. And since we are already given an explicit formula for y only in terms of x, it seems like we can just go ahead and take the derivative. But unfortunately, having an x both in the base and the exponent makes it a bit more complicated.

### So what can we do then?

Finding the derivative of this function is going to require a little trick that seems a little counter intuitive. What we need to do is actually take the natural log of both sides of this equation. The reason for this is that it will help us get rid of the exponent and put this in a form we can work with more easily.

$$y=x^x$$

$$ln (y) =ln \big( x^x \big)$$

Now that we have done this we can use one of the basic log rules that you will want to remember.

#### 3 log rules to remember

Really quickly I want to list the three main log rules that you will want to remember. These come up frequently, so you will want to remember these.

$$log(ab) = log(a) + log(b)$$ $$log \bigg( \frac{a}{b} \bigg) = lob(a) – log(b)$$ $$log \Big( a^b \Big) = b \cdot log(a)$$

#### How do we apply this to our problem?

Let’s look back to our equation to see where we were.

$$ln (y) =ln \big( x^x \big)$$

Notice the right side of our equation looks a lot like the third log rule from above. Based on that third log rule, we can move the x in the exponent down in front of the log and multiply rather than having to deal with an exponent.

$$ln(y) = x \cdot ln(x)$$

The reason I think this seems a little counter intuitive is that we no longer have an explicit formula for y. But now the right side of our equation will be easier to take its derivative. So now let’s see what happens if we take the derivative of both sides of the equation with respect to x.

### Taking the derivative

The reason we need to take the derivative with respect to x is that the question asked us to find $$\frac{dy}{dx}$$. The dx in the denominator is the indicator that tells us that we need to differentiate with respect to x. So let’s do that now.

$$\frac{d}{dx} ln(y) = \frac{d}{dx} \big[x \cdot ln(x) \big]$$

#### First the left side

Taking the derivative of the left side of the equation will require the use of the chain rule since y is a function of x. This was explained in a bit more detail in my implicit differentiation lesson. You will also use the fact that the derivative of ln(x) is $$\frac{1}{x}$$.

$$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx} \big[x \cdot ln(x) \big]$$

#### Then the right side

In order to take the derivative of the right side of this equation, we will need to use the product rule. As I did in the product rule lesson, we’ll first want to call one part of our function f and the other will be g.

$$f=x$$ $$g=ln(x)$$

Now we need to find f’ and g’ to use the product rule formula.

$$f’=1$$ $$g’=\frac{1}{x}$$

And lastly, we just need to plug these four pieces into the product rule formula.

$$\frac{1}{y} \cdot \frac{dy}{dx} = (x) \bigg( \frac{1}{x} \bigg) + (1) \big( ln(x) \big)$$

Now that we have taken the derivative of both sides of the equation, we just need to simplify the equation and solve for $$\frac{dy}{dx}$$.

### Solving for dy/dx

$$\frac{1}{y} \cdot \frac{dy}{dx} = (x) \bigg( \frac{1}{x} \bigg) + (1) \big( ln(x) \big)$$

$$\frac{1}{y} \cdot \frac{dy}{dx} = 1 + ln(x)$$

And all we need to do from here is multiply both sides by y to isolate $$\frac{dy}{dx}$$.

$$\frac{dy}{dx} = y \big( 1+ln(x) \big)$$

With most implicit differentiation problems this would be a perfectly fine place to stop and say we’ve reached our answer. Finding $$\frac{dy}{dx}$$ in terms of x and y is frequently the best we can do. But in this case, we can actually get our answer only in terms of x so that we have an explicit derivative of the original function.

The reason we’re able to do this is that our original function was an explicit formula for y. Since we know $$y=x^x$$ we can actually go to our formula for $$\frac{dy}{dx}$$ and replace the y with $$x^x$$. So,

$$\frac{dy}{dx} = x^x \big( 1+ln(x) \big)$$

And now we can say we have reached our answer!

I just want to circle back to those 3 log rules I discussed above. They can be very useful when taking derivatives of exponential functions, or in some strange cases products and quotients. They can be used to rewrite complex functions in a way that would make their derivative easier to find, so always be sure to be aware of those and know how to use them to manipulate a function when needed.

As I always say, the best way to learn this stuff is practice, practice, practice! So check out some of my other lessons and problem solutions on derivatives. The more you work with these concepts, the better you’ll start to understand them.

If you can’t find the answer to your question or the topic you want to read about on my site, send me an email at jakesmathlessons@gmail.com and I’ll get back to you as soon as I can. You can also use the form below to join my email list and I’ll send you my calculus 1 study guide!

## RELATED RATES – Sphere Volume Problem

The radius of a sphere is increasing at a rate of 4 $$\frac{mm}{s}$$. How fast is the volume increasing when the diameter is 80 mm?

This question states pretty clearly that we will be working with a sphere here. Since it gives information about how it is changing and asks us to find how quickly another value is changing, we know it’s a related rates problem.

As with all of the other related rates problems I’ve worked through, we are going to be following the same four step process. I will go through them one step at a time, but you can also find where I introduce the steps here.

## 1. Draw a sketch

The first thing we need to do is draw a sketch of the scene being described. Obviously we are dealing with a sphere, but we are really only told two things:

• how quickly the radius is changing, and
• what the diameter is at the specific moment we’re concerned with.

Since that’s all we know, it will be pretty simple to put that into a drawing. It would look something like this:

So we have a sphere whose radius is increasing at a rate of 4 $$\frac{mm}{s}$$ and we need to consider the moment when its diameter is 80 mm.

## 2. Come up with your equation

As with any related rates problem, we need to create our equation once we have created our drawing. To do this we need to consider the information we have been given and how it relates to the piece of information we’re looking for.

#### What are we looking for?

The question asks us to find how fast the volume is increasing when the diameter is 80 mm. Asking about how fast something is changing refers to its rate of change. Therefore, we can tell that this question is asking us about the rate of change of the volume.

Since we will later be taking the derivative of the equation we are currently building, we only need to make sure to include the volume. Once we take the derivative of the equation, this will introduce the rate of change of the volume.

#### What do we know about?

This question didn’t provide a lot of information and it’s fairly straight forward.

• Rate of change of the radius.
• The diameter of the figure at this moment.

#### Putting it into an equation

Up to this point we know that we need to include the sphere’s volume in this equation. We have also figured out that we know some information about the diameter, which can easily be used to find the radius. And we also know about the rate of change of the radius. So we basically know everything about the radius that we might need.

What is the first equation or formula you think of that relates the volume of a sphere to its radius?

I think a good place to start is with the formula for the volume of a sphere.

$$V=\frac{4}{3} \pi r^3$$

We can see that this equation only contains V (volume) and r (radius). As a result, I’d say this is as good of a place to start as any. Let’s proceed with this equation.

## 3. Implicit differentiation

Now that we have come up with our equation, we need to take its derivative with respect to time. This will allow us to introduce and work with the rates of change of our measurements.

Since we will be taking the derivative with respect to time, we will need to treat V and r as functions of time rather than variables. In order to do this we will need to use the chain rule. So, taking the derivative of our equation gives us:

$$\frac{d}{dt} \big[ V \big] = \frac{d}{dt} \bigg[ \frac{4}{3} \pi r^3 \bigg]$$

$$\frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt}$$

$$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}.$$

## 4. Solve for the desired rate of change

Finally, all we need to do now is solve for the rate of change the question is asking us to find. The problem asked us to find how fast the volume is changing at this moment. This is exactly what $$\frac{dV}{dt}$$ represents. Since this is already isolated, all we need to do is plug in the other values we know about.

The other values that are needed are r and $$\frac{dr}{dt}$$, which represent the radius and its rate of change respectively.

The radius of our figure wasn’t given directly, but we do know that its diameter is 80 mm at this instant. Since the radius of a sphere is always half of the diameter, this tells us that the radius is 40 mm, or

$$r=40.$$

We were given that the figure’s radius is increasing at a rate of 4 $$\frac{mm}{s}$$. Therefore, we know

$$\frac{dr}{dt} = 4.$$

#### Plugging it all in

Now we simply need to plug these values into the differentiated equation we found in step three.

$$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$$

$$\frac{dV}{dt} = 4 \pi (40)^2 (4)$$

$$\frac{dV}{dt} = 25,600 \pi$$

$$\frac{dV}{dt} \approx 80,424.772$$

So this tells us that the volume of the sphere is increasing at a rate of 25,600$$\mathbf{\pi \ \frac{mm^3}{s}}$$, or about 80,424.772 $$\mathbf{\frac{mm^3}{s}}$$ when its diameter is 80 mm.

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 practice problems that I have posted a solution of. 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!

I also encourage you to join my email list! Just enter your name and email below and I’ll send you a copy of my calculus 1 study guide to help you get through your homework faster and the exams with better scores!

## RELATED RATES – Cylinder Problem

A cylindrical tank with radius 5 m is being ﬁlled with water at a rate of 3 $$\frac{m^3}{min}$$. How fast is the height of the water increasing?

This is an interesting example because at first glance it doesn’t seem like we have been given enough information to solve this problem. If you compare this to the related rates cone problem we did, you can notice a few things that were given in that example but not this one.

• We don’t know the height of the cylindrical tank.
• We don’t know the height of the water at the instant we need to find its rate of change.

These things were both given in the cone problem we did.

You may want to check out this video of this problem. I solved it in a slightly different way in the video than I did in this post. Check out both so you can see it a couple different ways and decide which one makes more sense to you!

#### How do we deal with this missing information?

We actually don’t need it! Because the surface of the water in this tank will always be a circle with radius 5 m as this tank fills up, the height of the water or the tank won’t impact our answer. So there’s two possible ways to deal with this:

• Create an equation that doesn’t include the height of the tank or the water,
• or plug in some random value for the height and see what we get. If this one works we should be able to plug in any height and get the same answer for all of them.

One of these options actually won’t work, but we will explore them further later in this example. Let’s not get ahead of ourselves. Since we have a related rates problem here, we will want to follow the same four steps as all other related rates problems.

## 1. Draw a sketch

As with any related rates problem, the first thing we need to do is draw the situation being described to us. This is a relatively simple situation being described, so we can go ahead and draw it.

Looking at the above drawing, you can see that water is being poured into a cylindrical tank at a rate of 3 $$\frac{m^3}{min}$$. The height of the tank is unknown, but we know the radius is 5 m. We can also see that the water level is rising, but we don’t know the height of the water at this instant.

## 2. Come up with your equation

Now that we have a drawing of the situation being described, we need to come up with our equation. To do this, we need to sort out what information we know and what we are looking for.

#### What are we looking for?

The question is asking us to find how fast the height of the water is increasing. Once we take the derivative of our equation, that will introduce the variable representing “how fast things are changing.” Therefore, we know that we will need the height of the water to be in our equation.

Remember in the last section I said we may want to create an equation that didn’t contain the height of either cylinder because we have no information about them? Well now we know that’s not really an option. The question is asking us to find the rate of change of the height of one of the cylinders. To find this, we need the height of the cylinder to be in the equation.

#### What do we know about?

This question didn’t give us a lot of information but we can figure out a little extra. First let’s think about what was directly given. In order to do this, we should consider that this example essentially has two cylinders in it. The large cylinder is the tank, and the small cylinder is the water in the tank.

• We know that water is flowing into the tank at a rate of 3 $$\frac{m^3}{min}$$. This means that the volume of the small cone is increasing at a rate of 3 $$\mathbf{\frac{m^3}{min}}$$.
• The problem also says that the tank has a radius of 5 m.

And this is all the information that is explicitly given in the problem. However, there is one other fact we can infer. As the tank is filled with water, the liquid takes on a cylindrical shape as well. The cylinder made up of the water would be shorter than the tank, but it would have the same width. This leads to our last fact.

• We can conclude that the water also has a radius of 5 m.

#### Putting it into an equation

So far in this section, we have figured out that we will need to include the height of the water. We also determined that the information we know is about the radius of both cylinders and the rate of change of the volume of the small cylinder.

So we need our equation to relate the height, radius, and volume of the liquid. Because of this, I think a good place to start would be with the equation for the volume of a cylinder:

$$V=\pi r^2 h$$

In this equation V is the volume, r is the radius, and h is the height of the liquid in the tank.

One thing worth pointing out is that once we take the derivative of this equation, we will still have an h. But we don’t know anything about the height of the liquid in the tank at this instant. It seems like we wouldn’t have enough information to successfully use this equation, but it’s possible that the height at a given moment won’t actually impact the rate of change of the height. If this is true, we should be able to use this equation anyway. So let’s proceed and see what happens.

## 3. Implicit differentiation

As with any related rates problem, once we create our equation we need to take its derivative. Since we are taking the derivative with respect to time, we need to treat V, r, and h as functions of time rather than variables. Therefore, we need to use the chain rule. In this case, since we have r and h being multiplied, we will also have to use the product rule.

$$\frac{d}{dt}[V]=\frac{d}{dt} \big[\pi r^2 h \big]$$

### Using the product rule

To find the derivative of the right side of this equation we need to start by using the product rule. So we are trying to find

$$\frac{d}{dt} \big[\pi r^2 h \big].$$

Since $$\pi$$ is a constant being multiplied by the rest of the function we are taking the derivative of, we can pull it out of the derivative and deal with it later. Therefore, we can think of the right side of our equation as

$$\pi \frac{d}{dt} \big[r^2 h \big].$$

So we need to find the derivative of $$r^2h$$. Just as I did in the product rule lesson, we will start by deciding which part of this equation we will call f and g.

#### Choosing f and g

It doesn’t really make a difference which piece of this function we call f and which we call g. As long as we make a decision now and stick with it throughout our solution, it will work out in the end. We will say

$$f=r^2,$$

$$g=h.$$

#### Finding f’ and g’

Now that we have figured out f and g, the next step of the product rule is to find each of their derivatives. Keep in mind that we are taking the derivative with respect to time in this example.

To find f’, we will need to use the chain rule as well since r is a function of time. doing this tells us that

$$f’=2r \frac{dr}{dt}$$

Lastly, we need to find g’ by taking the derivative of g with respect to time.

$$g’=\frac{dh}{dt}$$

#### Putting the pieces of the product rule together

Now we have figured out all four of the pieces of the product rule, so we can just plug them into the product rule formula to find the derivative we’re looking for.

$$\frac{d}{dt} \big[r^2 h \big] = 2r \frac{dr}{dt} \cdot h + \frac{dh}{dt} \cdot r^2$$

And lastly, we just need to bring the $$\pi$$ back into the equation.

$$\frac{d}{dt} \big[\pi r^2 h \big] = \pi \bigg[ 2r \frac{dr}{dt} \cdot h + \frac{dh}{dt} \cdot r^2 \bigg]$$

### Putting it all together

Now that we have the derivative of the right side of our equation, we can go back and figure out the left side.

$$\frac{dV}{dt}= \pi \bigg[ 2r \frac{dr}{dt}h + \frac{dh}{dt}r^2 \bigg]$$

## 4. Solve for desired rate of change

The last step of any related rates problem is to solve for the rate of change we need to find. The question asks us to find how fast the height of the water is increasing. This is exactly what $$\frac{dh}{dt}$$ represents, so we need to isolate that variable.

$$\frac{dV}{dt}= \pi \bigg[ 2r \frac{dr}{dt}h + \frac{dh}{dt}r^2 \bigg]$$

$$\frac{1}{\pi}\frac{dV}{dt}= 2r \frac{dr}{dt}h + \frac{dh}{dt}r^2$$

$$\frac{1}{\pi}\frac{dV}{dt} – 2r \frac{dr}{dt}h = \frac{dh}{dt}r^2$$

$$\frac{1}{r^2} \bigg[ \frac{1}{\pi}\frac{dV}{dt} – 2r \frac{dr}{dt}h \bigg] = \frac{dh}{dt}$$

Now that we have isolated the term we need to solve for, we just need to plug in the values for all of the other variables. Before we just into this, I want to focus on just one of these variables.

#### What do we do with h?

Remember back in the beginning of this solution I said we may need to make up some value for h and just go with it? Let’s consider this for a moment.

Notice that h only appears in our equation in one place and it is being multiplied by two other variables, r and $$\frac{dr}{dt}$$. Think about what $$\frac{dr}{dt}$$ represents. It is the rate of change of the radius of the water. But our water has a constant radius, it’s always 5 m.

Since the radius of the cylinder is never changing, its rate of change must always be zero! Therefore, we know that

$$\frac{dr}{dt} = 0.$$

Since h is being multiplied by another term that is always zero, it’s not going to matter what h is. We can essentially use any number for h and it won’t matter because we are going to multiply it by zero anyways. We can also just leave it as h and not plug anything in for it.

#### Back to plugging in our values

We have already figured out that $$\frac{dr}{dt}=0$$ and luckily the other variables’ values were given.

Looking back at the original sketch, you can see that the radius of our cylinder is 5 m. So

$$r=5.$$

Also, you can see that water is flowing into the tank at a rate of 3 $$\frac{m^3}{min}$$. Since this is the only factor that will be impacting the volume of water in the tank, this must be the exact rate that the volume of the water is increasing.

$$\frac{dV}{dt}=3$$

Now we can plug all of these into the equation then simplifying to get our answer!

$$\frac{1}{r^2} \bigg[ \frac{1}{\pi}\frac{dV}{dt} – 2r \frac{dr}{dt}h \bigg] = \frac{dh}{dt}$$

$$\frac{1}{(5)^2} \bigg[ \frac{1}{\pi}(3) – 2(5)(0)h \bigg] = \frac{dh}{dt}$$

$$\frac{1}{25} \bigg[ \frac{3}{\pi} – 0 \bigg] = \frac{dh}{dt}$$

$$\frac{1}{25} \cdot \frac{3}{\pi} = \frac{dh}{dt}$$

$$\frac{3}{25 \pi} = \frac{dh}{dt}$$

So we now know that the rate of change of the height of the water is $$\frac{3}{25 \pi}$$ no matter what the height of the cylinder is at that moment. In other words, we can say that the height of the water is increasing at a rate of

$$\mathbf{\frac{3}{25 \pi} \ \frac{m}{min}}.$$

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 a solution of. 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!

Also, if you want a copy of my calculus 1 study guide, just enter your name and email below and I’ll send you a copy and add you to my email list so you can see as soon as I have more knowledge to share!

## RELATED RATES – Sphere Surface Area Problem

If a snowball melts so that its surface area decreases at a rate of 1 $$\frac{cm^2}{min}$$, find the rate at which the diameter decreases when the diameter is 10 cm.

By looking at the given statement, we can gather a few important fact quickly.

• The object in question is a snowball. This means that we will be dealing with a sphere here, so we will likely be needing some formula relating different dimensions and measurements of a sphere.
• The problem gives information about the rate of change of a specific measurement of the snowball. When a problem gives information like this, it’s a strong hint that we have a related rates problem.

So we know that we’re dealing with a related rates problem. Therefore, we are going to follow the four steps that these will all follow. If you want to look back at these steps, I discussed them in my related rates lesson.

## 1. Draw a sketch

Looking back at the problem, you can see that there isn’t a lot of information that has been described to us. This will end up being a very simple sketch, but that’s all it takes sometimes.

So we have a sphere whose surface area is decreasing at a rate of 1 $$\frac{cm^2}{min}$$ and we are looking at the instant when its diameter is 10 cm.

## 2. Come up with your equation

Now that we have our situation drawn out, we need our equation. To create this equation, we need to incorporate any relevant information that we were given and we need to consider what the question asks us to find.

#### What are we looking for?

The question is asking us to find “the rate at which the diameter decreases” at the instant when the diameter is 10 cm. This is the instant that we captured in our drawing. So the important thing here is that we are looking for the rate the diameter is decreasing. Another way to put this is the rate of change of the diameter.

Since we will eventually need to take the derivative, which will provide the rate of change part, we just need to make sure that our equation contains the diameter.

#### What do we know about?

This question didn’t provide a lot of information to us. We really only know two things:

• The rate of change of the sphere’s surface area.
• The diameter of the sphere at this instant.

#### Putting it into an equation

Up to this point we have figured out that we need to include the sphere’s diameter in our equation. We also know that we have some information relating to the snowball’s surface area. So we need to come up with an equation that relates a sphere’s surface area and diameter.

A good place to start may be the equation for the surface area of a sphere.

$$A=4 \pi r^2$$

In this equation A represents the surface area of the sphere and r is the radius.

Since we need an equation relating the surface area and the diameter, we will need to make an adjustment. The only thing we need to consider here is that a sphere’s diameter is always double the radius. Or in other words

$$d=2r.$$

But our equation contains the radius. So we’ll want to solve for r so we can plug that into our equation. Dividing both sides by 2 will accomplish this.

$$r=\frac{d}{2}$$

Now we can plug this in for r in our equation for the surface area of a sphere.

$$A=4 \pi \bigg( \frac{d}{2} \bigg)^2$$

So we know have an equation that contains only the surface area and the diameter of our sphere, exactly what we needed. To make things a little easier later, we will simplify this equation a bit.

$$A=4 \pi \bigg( \frac{d^2}{4} \bigg)$$

$$A=\pi d^2$$

## 3. Implicit differentiation

Now that we have created our equation we need to take its derivative. This will bring in the rates of change we discussed earlier. The most important one being the rate of change of the snowball’s diameter which is what we need to find.

Keep in mind that we will be taking the derivative with respect to time. This means we will need to treat A and d as functions of time, not variables. Therefore, we will need to apply the chain rule here.

$$\frac{d}{dt} \big[ A \big]= \frac{d}{dt} \big[ \pi d^2 \big]$$

$$\frac{dA}{dt} = 2 \pi d \cdot \frac{dd}{dt}$$

## 4. Solve for the desired rate of change

The fourth and final step of a problem like this is to isolate the rate of change we need and find its value. The question wants us to find the rate at which the diameter is decreasing when the diameter is 10 cm. This tells us we need to solve for the rate of change of the diameter, which is represented by $$\frac{dd}{dt}$$.

$$\frac{dA}{dt} = 2 \pi d \cdot \frac{dd}{dt}$$

$$\frac{1}{2 \pi d} \frac{dA}{dt} = \frac{dd}{dt}$$

So now that we have isolated the variable that we need to find we can simply plug in all of the values we have for the other variables.

#### What values do we know?

There are only two variables we need to plug in a value for: d and $$\frac{dA}{dt}$$. The question specifically asked us to find $$\frac{dd}{dt}$$ when the diameter of the snowball is 10 cm. Therefore, we know that we will have

$$d=10.$$

The question also told us that the surface area of the snowball is decreasing at a rate of 1 $$\mathbf{\frac{cm^2}{min}}$$. Since the surface area is decreasing, this tells us that

$$\frac{dA}{dt}=-1.$$

#### Plugging it all in

Now we just need to go back to our equation and plug in all of these other values.

$$\frac{dd}{dt} = \frac{1}{2 \pi d} \frac{dA}{dt}$$

$$\frac{dd}{dt} = \frac{1}{2 \pi (10)} \cdot (-1)$$

$$\frac{dd}{dt} = \frac{-1}{20 \pi}$$

So this tells us that the diameter of the snowball is changing at a rate of $$\frac{-1}{20 \pi} \ \frac{cm}{min}$$. Therefore, we can say that the diameter of the snowball is decreasing at a rate of $$\mathbf{\frac{1}{20 \pi} \ \frac{cm}{min}}$$ or about 0.0159 $$\mathbf{\frac{cm}{min}}$$.

Note that the question asked for the rate at which the diameter is decreasing. As a result of this, our answer will actually be a positive number despite the fact that $$\mathbf{\frac{dd}{dt}}$$ was negative.

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 these types of problems that I have posted a solution of. 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!

Enter your name and email below and I’ll send you a FREE copy of my calculus 1 study guide as well! It’s packed full of helpful formulas and shortcuts to help you get through calculus easier!

## RELATED RATES – Cone Problem (Water Filling and Leaking)

Water is leaking out of an inverted conical tank at a rate of 10,000 $$\frac{cm^3}{min}$$ at the same time water is being pumped into the tank at a constant rate. The tank has a height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 $$\frac{cm}{min}$$ when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

Here we have another related rates problem. This is a pretty typical problem you would see in a calculus class. There is a lot going on in this one since we have a related rates with a cone filling and leaking water. Just like I said when I discussed related rates, these problems tend to follow a specific pattern. If you need a refresher on what the four steps are just click that link to my related rates lesson. Otherwise, we’ll jump right into it.

## 1. Draw a sketch

You should always start a related rates problem with a drawing of the real world situation that’s being described in the problem. The problem describes an “inverted conical tank.” This just means that the tank is in the shape of an up-side-down cone. Other than that, the other facts are quite simple.

• Water is leaking out at a rate of 10,000 $$\frac{cm^3}{min}$$.
• Liquid is being pumped into the tank at an unknown constant rate.
• Tank’s height is 6 m.
• Diameter of the circular opening is 4 m.
• Water level is rising at a rate of 20 $$\frac{cm}{min}$$ when the height of the water is 2 m.

So we need to put all of these facts into a drawing, which might look something like this.

The important thing to point out about our sketch is that we have two cones here.

One cone is the tank, which is the larger cone. This one will always stay the same size and is not changing. This means that its height, diameter of the base, and its volume are all constants.

The second cone is the water sitting in the bottom of the tank, which is the smaller cone. This one is changing as our liquid flows into and out of the tank. As time passes its dimensions change. Therefore, its height, diameter of the base, and volume are all functions of time.

## 2. Come up with your equation

Now that we have made our sketch, we need to come up with an equation that relates all of the different quantities we’re dealing with. All of the information we know about and the information we are looking for relates to a volume or measurements of some cone. The measurements are either of our water in the tank or the tank itself, but in both cases the measurements describe a cone.

#### What are we looking for?

The question asks us to find the rate at which water is being pumped into the tank. As it is pumped into the tank, this will impact the volume of the smaller cone which is the water sitting in the tank. Therefore, the information we are looking for will somehow relate to how quickly the volume of the liquid in the tank is changing. Or the rate of change of the volume of the small cone.

#### What do we know about?

We were given quite a bit of information so let’s break it into a few different pieces.

• We know the height and base diameter of the tank. Using these, we can find its base radius and volume. Since the tank isn’t changing we know, or can easily find, any important measurement of the large cone.
• We know how fast water is flowing out of the tank. This tells us a piece of information relating to the rate of change of the small cone. We will need to use this once we know how fast water is flowing into the tank.
• The last information we already know is the height and the rate of change of the height of the small cone at a particular moment.

#### Putting it into an equation

As I mentioned above, we need to find the rate of change of the volume of the liquid in the tank. Since we know we will need to use implicit differentiation to get the rate of change, our equation needs to involve the volume of the small cone. Remember, the equation we come up with should include quantities and measurements, not rates of change yet.

So we are dealing with the volume of something that’s in the shape of a cone. Because of this we can start by looking at the formula for a volume of a cone then we’ll see if we can use this directly or if we will need to make some adjustments. The volume of a cone can be described by

$$V=\pi r^2 \frac{h}{3}$$

where r is the radius of the cone’s base and h is its height.

Clearly we would be able to use this to get the rate of change of the volume since the equation already includes the volume. But this means we need to know everything else in the equation.

Other than the volume, the equation requires that we know the radius and height of the cone. We also need to know the rate of change of these other parts.

By referring back to our original drawing, you can see that we clearly already know the height and the rate of change of the height of the liquid in the tank at this specific moment. Remember, these two values are changing as liquid goes into the tank, but we know their values at this specific instant.

We don’t know the radius yet, but we can find it by comparing the small and large cones and using similar triangles. However, the problem is that we don’t know the rate of change of the radius. We could figure it out, but it would be quite complicated and there is actually something else we can do instead.

So we want to avoid using the radius, or diameter, because we don’t want to have to find its rate of change. So what else can we do?

In order to get around this, we will need to write an equation that doesn’t include the radius. But remember, it’s fine if we use the height because we know its rate of change. The important fact we need to use here is that the two cones will always form similar triangles. This will be true no matter where the water level is, as it fills up.

Consider the following drawing which represents the conical tank shown from the side.

As shown in the sketch above, the top sides of these triangles are parallel. This means that each angle in the small triangle is the same as the corresponding angle in the large triangle. Therefore, these are similar triangles. Since they are similar triangles, we know one important fact:

$$\frac{x}{2}=\frac{4}{6}$$

So we can use this to solve for x.

$${x}=\frac{4}{6} \cdot 2$$

$$x=\frac{4}{3}$$

So we can see that in both of these triangles, the top side is two-thirds as long as the height of the triangle. In other words, if we multiply the height of the triangle by $$\frac{2}{3}$$, this would give us the length of the top side of the triangle. Remember the top of the triangle is the diameter of the corresponding cone from our first drawing. This tells us the following relationship between the height and diameter of the cones:

$$\frac{2}{3}h=d$$

Since our goal here is to find out the relationship between the radius and the height, not the diameter and the height, we need to do one more thing. The diameter of a circle is always twice as much as the radius, so we know $$d=2r$$. We can substitute this in for d and solve for r.

$$\frac{2}{3}h=2r$$

$$\frac{1}{3}h=r$$

#### Going back to our equation

Now we need to go all the way back to our volume equation that involved h and r.

$$V=\pi r^2 \frac{h}{3}$$

Since we now have this new way to write r in terms of h, we can substitute this into the volume equation and simplify from there.

$$V=\pi \bigg( \frac{1}{3}h\bigg)^2 \frac{h}{3}$$

$$V=\pi \cdot \frac{1}{9}h^2 \frac{h}{3}$$

$$V= \frac{1}{27} \pi h^3$$

Now we have an equation involving V and h. Since we are looking for the rate of change of V, and we already know about V, h, and the rate of change of h, this equation will work perfectly.

## 3. Implicit differentiation

We created our equation containing all of the necessary pieces. Remember earlier I said that we are going to need to find the value of $$\frac{dV}{dt}$$ in order to find how quickly liquid is going into the tank? Now is the time to do this.

As with any related rates problem, we need to take the derivative of our equation. Since we already have an equation for V, all we need to do to find $$\frac{dV}{dt}$$ is take the derivative of V with respect to time. Let’s do that now.

$$\frac{d}{dt} \big[ V \big] = \frac{d}{dt} \bigg[ \frac{1}{27} \pi h^3 \bigg]$$

The left side of this equation will be quite simple, but to find the right side we need to keep a couple things in mind. We are going to take the derivative with respect to time. Therefore, we need to treat h as a function of time rather than a variable. This means that we will need to use the chain rule to take this derivative. You can see a more detailed explanation of how to do this and why we are doing it in my implicit differentiation lesson.

Also remember that $$\frac{1}{27} \pi$$ is just a constant coefficient in front of our $$h^3$$ term.

$$\frac{dV}{dt}= \frac{3}{27} \pi h^2 \cdot \frac{dh}{dt}$$

$$\frac{dV}{dt}= \frac{1}{9} \pi h^2 \cdot \frac{dh}{dt}$$

## 4. Solve for desired rate of change

The final step of a related rates problem is to solve of the rate of change the question asked for. Since $$\frac{dV}{dt}$$ is already isolated we don’t need to worry about solving for the variable we are looking for.

$$\frac{dV}{dt}= \frac{1}{9} \pi h^2 \cdot \frac{dh}{dt}$$

All we need to do now is plug in the other information missing from the equation to tell us all we need to know about $$\frac{dV}{dt}$$. As you can see, this mean we need to figure out the values for h and $$\frac{dh}{dt}$$. Luckily, both the height and the rate the height is increasing of the small cone were both given to us.

Looking back at our drawing, which I have put below, we can see that both of these values we need are already labeled.

Looking at our sketch we can see that h, which is shown as the height of our small cone, is 2 m. We can also see that $$\frac{dh}{dt}$$, which is shown as the speed at which the water level is rising, is 20 $$\frac{cm}{min}$$. But there is one problem here. We need to be careful of our units!

Notice that the height is measured in meters and its rate of change is measured in centimeters.

We need to convert one of these two measurements so that we are either using meters or centimeters throughout the whole problem. It doesn’t matter which one we choose as long as there are the same. I will go ahead and use meters. Don’t worry, if you need the answer in centimeters I’ll discuss how to convert to that in a minute. Since $$100 \ cm = 1 \ m$$ we know that

$$h=2m$$

$$\frac{dh}{dt} = 0.2 \ \frac{m}{min}$$

#### Putting it all together

So now we can plug these values back into our equation for $$\frac{dV}{dt}$$.

$$\frac{dV}{dt}= \frac{1}{9} \pi h^2 \cdot \frac{dh}{dt}$$

$$\frac{dV}{dt}= \frac{1}{9} \pi (2)^2 \cdot 0.2$$

$$\frac{dV}{dt}= \frac{4}{9} \pi \cdot \frac{1}{5}$$

$$\frac{dV}{dt}= \frac{4 \pi}{45}$$

$$\frac{dV}{dt} \approx 0.279 \ \frac{m^3}{min}$$

But remember this isn’t quite our answer. We still have one more step. The question asked us to find the rate at which water is being pumped into the tank. We just found the rate at which the volume of water in the tank is changing.

There are three important values to think about here.

• The rate water is being pumped into the tank.
• Rate of change of the water actually in the tank.
• The rate liquid is flowing out of the tank.

We are trying to find the rate water is being pumped into the tank and we already know the other two rates. It was given that water is flowing out of the tank at a rate of 10,000 $$\frac{cm^3}{min}$$.

#### Be careful of the units

We know the rate at which the liquid is flowing out of the tank, but it uses centimeters. Since we decided earlier that we want to use meters instead, we need to convert this as well. This will be a little different though.

Think about a cube with each side length being 1 m. This cube would be 1m x 1m x 1m and would have a volume of 1 $$m^3$$. Since we know that $$1 \ m = 100 \ cm$$, we can also say that this same cube is 100cm x 100cm x 100cm and would actually have a volume of 1,000,000 $$cm^3$$. Therefore, we can say

$$1m^3 = 1,000,000cm^3.$$

If we divide both sides of this equation by 100, we can see that

$$\frac{1}{100}m^3 = 10,000cm^3.$$

So we then can say that water is flowing out of the tank at a rate of 0.01 $$\frac{m^3}{min}$$.

#### How it fits together

We know that $$\frac{dV}{dt} \approx 0.279 \ \frac{m^3}{min}$$ is a positive number. This means that the amount of water in the tank is increasing at this instant. Since some water is flowing out of the tank at a rate of 0.01 $$\frac{m^3}{min}$$, it must be flowing into the tank faster than this. If it were not, the volume in the tank would be decreasing.

In fact, you can think of the rate of change of the volume of water in the tank as the rate of water flowing in minus the rate of water flowing out.  Let’s say X represents the rate at which liquid is flowing into the tank and we can use this equation to represent the previous sentence.

$$X-0.01=0.279.$$

Solving for X tells us

$$X=0.289.$$

So water must be flowing into the tank at a rate of 0.289 $$\frac{m^3}{min}$$. And this is the value the question was asking us to find!

#### We can also give our answer in other units

If we go back to our equation before we started plugging in anything, we can instead plug everything in terms of centimeters instead of meters. In this case we will have $$h=200cm$$ and $$\frac{dh}{dt}=20 \frac{cm}{min}$$.

$$\frac{dV}{dt}= \frac{1}{9} \pi h^2 \cdot \frac{dh}{dt}$$

$$\frac{dV}{dt}= \frac{1}{9} \pi (200)^2 (20)$$

$$\frac{dV}{dt} \approx 279,252.68$$

Now just like above, we need to also consider how fast the water is flowing out of the tank. This time we don’t need to convert the units though since we are using centimeters.

$$Y-10,000= 279,252.68$$

$$Y= 289,252.68$$

So we can also say that water is flowing into the tank at a rate of 289,252.68 $$\frac{cm^3}{min}$$. This is more likely going to be a better answer simply because it is more precise that the previous answer we found!

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This was a good cone related rates example, but if you 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 and 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!

## 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?

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?

### 1. Draw a sketch

Here we have a related rates problem.  As I said when I discussed related rates problems initially, the first thing I like to do with these problems is draw a sketch of the scene that is being described.  If you want to refer back to that, you can click here.  Otherwise, let’s sketch the problem described here.

### 2. Come up with your equation

The next thing we need to do is set up our equation which will relate our different quantities.  To do this, we will want to consider what value the question is asking us to find.

What are we looking for?

It asks “at what rate is the angle between the string and the horizontal decreasing when 200 ft of string has been let out?”  Therefore, the value we are looking for is the “rate of change of the angle between the string and the horizontal.”  This just means we will need to consider the angle between the string and the ground (the ground is the horizontal in this case).  If you look back at our drawing, you will see that this angle is represented by $$\theta$$.  Since our goal is to find how fast $$\theta$$ is changing, we need $$\theta$$ to be in our original equation.

Now we need to consider what other quantities or variables we know something about.  Clearly we know something about the two sides of the triangle that are labeled as being 100 ft and 200 ft.  And we can use these two sides to figure out the length of the third side, which is not labeled in our drawing.

Although we could simply call one of those sides $$a$$ and the other one $$b$$ and proceed from there, there is another option that may simplify our problem.

Consider the fact that the kite is moving horizontally.  This means that the kite is not getting any further from or closer to the ground as it moves.  Therefore, the side that is labeled 100 ft will actually be 100 ft at any point in this kite’s flight.  Because of this we actually don’t need to designate a variable to this side of the triangle.  Instead this side is simply a constant 100 ft.

Now we just need to use one of the other two sides of the triangle.  We could technically use either one, but one will be a lot easier than the other.  It looks like the hypotenuse would be the easier of the two, because we know it’s 200 ft at this moment.  However, we don’t know exactly how fast it’s changing.  We can figure that out but it wouldn’t be easy.

We do know exactly how fast the unlabeled side is changing.  The question states that the kite is moving horizontally at a speed of 8 $$\frac{ft}{s}$$.  Since this unlabeled side is exactly horizontal, we know its rate of change is also 8 $$\frac{ft}{s}$$.  We can figure out its length using Pythagorean Theorem later, but this would certainly be easier than finding the rate of change of the hypotenuse.  Therefore, I will go ahead and use the unlabeled side.

Since this unlabeled side is going to be changing we will need to designate a variable to this side of the triangle.  As the kite moves away from the person flying it, the person holding the string has to let more string out and allow it to become longer.  This means that this unlabeled side in our drawing will need to be described with a variable.  We will call it side $$a$$.

Putting it into an equation.

Now we have three different quantities we need to relate somehow:

1. Angle $$\theta$$ (this will be changing as the kite moves).
2. Side $$a$$ (this will be changing as the kite moves and the string is let out).
3. Side labeled 100 ft (this will not change and can be treated as a constant).

So we have two sides and an angle that we need to make an equation with.  To do this, think about where these sides are in relation to the angle $$\theta$$.  The side labeled 100 ft is the side opposite to the angle $$\theta$$ and the side we’re calling $$a$$ is adjacent to the angel $$\theta$$.

Usually when dealing with two sides and one angle of a triangle, you will want to use either sine, cosine, or tangent to relate the three.  So which one should be used when we know the opposite side and the adjacent side to the angle in question?

Remember soh, cah, toa?

• Sine Opposite Hypotenuse

Since we have the opposite side and the adjacent side, we want to use tangent.  Therefore we can say:

$$tan(\theta) = \frac{100}{a}$$

Since it will make finding the derivative easier, I am going to rewrite this as

$$tan(\theta) =100a^{-1}$$

### 3. Implicit differentiation

As with any related rates problem, we now need to take the derivative of both sides of the equation with respect to time.  Since $$\theta$$ and $$a$$ are both functions of time, we will need to use chain rule for both sides of this equation.  We know they are functions of time because they are both going to be dependent on the position of the kite as time progresses.  We don’t have an explicit formula for either of these functions, but we know their values are dependent on time.

$$\frac{d}{dt}tan(\theta) =\frac{d}{dt}100a^{-1}$$

$$\frac{d}{dt}\frac{sin(\theta)}{cos(\theta)} =\frac{d}{dt}100a^{-1}$$

To find the derivative of the left side of this equation you will need to use the quotient rule and the chain rule.  I’m not going to show all the steps of how to do this but if you want a refresher, you can read about the quotient rule here and the chain rule here.  Using Wolfram Alpha, you can see that

$$\frac{d}{dx}tan(x)=\frac{1}{cos^2x}$$

Therefore, we can say that

$$\frac{d}{dt}tan(\theta)=\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt}$$

Plugging this back into the left side of our equation, we get

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =\frac{d}{dt}100a^{-1}$$

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt}$$

### 4. Solve for desired rate of change

The last step of any related rates problem is to solve for the desired rate of change.  Now remember the thing we need to find is the rate of change of our angle $$\theta$$.  This is exactly what $$\frac{d\theta}{dt}$$ represents.  So now we just need to solve for $$\frac{d\theta}{dt}$$.

$$\frac{1}{cos^2 \theta} \cdot \frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt}$$

$$\frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt} \cdot cos^2 \theta$$

Now we just need to plug in the values for $$a$$, $$\frac{da}{dt}$$, and $$\theta$$ and we will have our answer.  We don’t know all of these values but we can find them.

Finding a

As I mentioned before, we can find $$a$$ by using Pythagorean Theorem.  Looking back at our drawing, we have a right triangle with side lengths of 100 ft, 200 ft, and $$a$$.  We know that

$$100^2 + a^2 = 200^2$$

$$10,000 + a^2 = 40,000$$

$$a^2 = 30,000$$

$$a = \sqrt{30,000}$$

$$a = 100\sqrt{3}$$

Finding  $$\mathbf{\frac{da}{dt}}$$

This was actually given.  We know that $$a$$ is the horizontal distance the kite is away from the person flying the kite.  We know that the kite is moving horizontally at a speed of 8 $$\frac{ft}{s}$$.  Because of this we know that this is also the rate at which $$a$$ is changing.  Since $$\frac{da}{dt}$$ is the rate of change of $$a$$, we know

$$\frac{da}{dt} = 8$$

Finding $$\mathbf{\theta}$$

To find $$\theta$$ we will need to go back to the original equation we came up with before the implicit differentiation step:

$$tan(\theta) = \frac{100}{a}$$

Since we know $$a$$, we can plug it in here and solve for $$\theta$$.

$$tan(\theta) = \frac{100}{100\sqrt{3}}$$

$$tan(\theta) = \frac{1}{\sqrt{3}}$$

This angle is actually on the unit circle and by using this we know:

$$\theta = \frac{\pi}{6}$$

Note that $$\theta$$ will be in radians.

Now we can plug all of these into our equation for $$\frac{d\theta}{dt}$$.

$$\frac{d\theta}{dt} =-100a^{-2} \cdot \frac{da}{dt} \cdot cos^2 \theta$$

$$\frac{d\theta}{dt} =-100 \big(100\sqrt{3} \big)^{-2} \cdot 8 \cdot cos^2 \Bigg( \frac{\pi}{6} \Bigg)$$

$$\frac{d\theta}{dt} =-\frac{1}{300} \cdot 8 \cdot \Bigg( \frac{\sqrt{3}}{2} \Bigg)^2$$

$$\frac{d\theta}{dt} =-\frac{1}{300} \cdot 8 \cdot \frac{3}{4}$$

$$\frac{d\theta}{dt} =-\frac{1}{50}$$

So we can say that the angle between the string and the horizontal is decreasing at a rate of $$\frac{1}{50} \ \frac{radians}{s}$$ when 200 ft of string has been let out.

And that’s the answer to the question!  Hopefully that wasn’t too bad, but if you have any questions I’d love to hear them.  I know related rates problems can be challenging so you can email me any questions or suggestions at jakesmathlessons@gmail.com.  If you have any other problems you’d like to see worked out go ahead and send me an email.

If you feel you need some more practice with related rates, you can check out the lesson where I discussed related rates for more examples.

Also, if you want to check out some other problems and get some practice with derivatives, go check out my derivatives page.  You can see what other topics I’ve already covered and problems I’ve worked through.  If you can’t find your problem there just let me know and I may post the solution to your problem.

## Solution – Find the values of a and b that make the function differentiable everywhere.

Find all values of $$a$$ and $$b$$ that make the following function differentiable for all values of $$x$$.

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ ax+b & \mbox{if } x>-1 \end{cases}$$

When trying to solve a problem like this, there are actually two things you will need to consider for our function $$f(x)$$.  Obviously, we need to make sure that it’s differentiable everywhere, but this actually implies something else that we will want to consider as well.

Since a function being differentiable implies that it is also continuous, we also want to show that it is continuous.  The reason for this is that any function that is not continuous everywhere cannot be differentiable everywhere.  Once we make sure it’s continuous, then we can worry about whether it’s also differentiable.

## Making sure f(x) is continuous everywhere

I’m not going to go into quite as much detail to show the part about making sure the function is continuous because I have already done this, which you can see by clicking here.

To make sure $$f(x)$$ is continuous at $$x=-1$$ we need to make sure that $$\lim_{x \to -1} f(x) = f(-1).$$  Since we have a piecewise function, we will need to consider each one-sided limit, but in this case only the right sided limit will tell us something useful.

$$\lim_{x \to -1^{+}} f(x) = f(-1)$$

$$\lim_{x \to -1^{+}} ax+b = b(-1)^2-3$$

$$a(-1)+b=b-3$$

$$-a+b=b-3$$

$$-a=-3$$

$$a=3$$

So now we know that $$f(x)$$ will be continuous everywhere as long as $$a=3$$.  However, this doesn’t really tell us that $$f(x)$$ is differentiable everywhere as well.

## Making sure f(x) is differentiable everywhere

We now know that we will need to let $$a=3$$ in order for this function to be continuous and to have a chance of being differentiable.  As a result, we can say that we are now trying to make this function differentiable everywhere:

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ 3x+b & \mbox{if } x>-1 \end{cases}$$

We can see that the only place this function would possibly not be differentiable would be at $$x=-1$$.  The reason for this is that each function that makes up this piecewise function is a polynomial and is therefore continuous and differentiable on its entire domain.  The only place we may have a problem is when we have to switch between the two functions.

#### What does it mean for a function to be differentiable?

It means that its derivative exists for all values of $$x$$.  In other words, we need to be able to find its derivative no matter what $$x$$ is.

However, as I mentioned above, in this case we really only need to make sure that we can find the derivative of $$f(x)$$ when $$x=-1$$ since we know it would exist for all other values of $$x$$.  By using the definition of a derivative, we need to make sure the following limit exists at $$x=-1$$.

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Since we need to check this when $$x=-1$$, we can plug in $$-1$$ for $$x$$.  Therefore, we need to make sure this limit exists:

$$\lim_{h \to 0} \frac{f(-1+h)-f(-1)}{h}$$

I went over this limit definition in greater detail previously.  If you want a refresher on where this is coming from you can find that by clicking here.

Just like when we had to find the limit to make sure that $$f(x)$$ was continuous, we will need to consider each one sided limit separately in order to find this limit.  And also like when we checked for continuity, each one sided limit is going to require the use of a different section of our piecewise function.

#### Setting up the limits

When $$h$$ is slightly less than $$0$$, and we are considering the left sided limit, $$f(-1+h)$$ would need to be found using the $$y=bx^2-3$$ because this would involve inputting $$x$$ values which are less than $$-1$$.

By the same reasoning, when $$h$$ is slightly greater than $$0$$, and we are considering the right sided limit, $$f(-1+h)$$ would need to be found using the $$y=3x+b$$ because this would involve inputting $$x$$ values which are greater than $$-1$$.  Therefore, we need to consider the following one sided limits:

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

#### Now what do we do with these limits?

Now remember, as I discussed in the lesson about one-sided limits, in order for a limit to exist we need both of its one-sided limits to exist and they need to be equal.  Therefore, in order to show that the derivative of $$f(x)$$ exists at $$x=-1$$, these two one-sided limits need to be equal to each other.  Before setting them equal to each other, first we’ll simplify them a bit.  First the left side limit.

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)(-1+h)-3\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(1-2h+h^2)-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b-2bh+bh^2-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{b-2bh+bh^2-3-b+3}{h}$$

$$=\lim_{h \to 0^{-}} \frac{bh^2-2bh}{h}$$

$$=\lim_{h \to 0^{-}} \frac{h(bh-2b)}{h}$$

$$=\lim_{h \to 0^{-}} bh-2b$$

$$=-2b$$

And now the right sided limit.

$$=\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{\Big[-3+3h+b\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{-3+3h+b-b+3}{h}$$

$$=\lim_{h \to 0^{+}} \frac{3h}{h}$$

$$=\lim_{h \to 0^{+}} 3$$

$$=3$$

Now if we set these two simplified versions of the one-sided limits equal to each other, we get

$$-2b=3$$

$$b=-\frac{3}{2}$$

#### What does this tell us?

So now if we put both pieces together, we know that $$a=3$$ will ensure that $$f(x)$$ is continuous and then making $$b=-\frac{3}{2}$$ will also make sure $$f(x)$$ is differentiable at $$x=-1$$.  This would in turn make $$f(x)$$ differentiable for all values of $$x$$, or make it differentiable everywhere.

As always, I want to hear your questions!  Go check out my other lessons about derivatives and if you can’t get your question answered, I’d love to hear from you.  Leave a comment below or email me at jakesmathlessons@gmail.com.  If you have questions on this problem and solution or if you have another question you would like to see me answer, just ask it.  Or if you have an entire topic you would like to see me write a lesson about, just let me know.

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## Solution – Find two numbers whose sum is 23 and whose product is a maximum

Find two numbers whose sum is 23 and whose product is a maximum.

When you see a problem like this, it’s obvious our goal is to maximize some function.  The more challenging part is figuring out what the function would be that we want to maximize.  When we look at this problem however, we can actually break it down into two different facts we know about these two numbers, which we will call a and b.

• The two numbers add up to 23.
• We need to maximize their product.

Each of these two bullet points can actually be represented mathematically and they can be used together.

#### Fact #1

First of all, we know that these two numbers add up to 23.  Or in other words,

$$a+b=23.$$

#### Fact #2

Also, we need to maximize the product of these numbers.  Another way to put this is that we want to maximize the function f, which represents the product of a and b.  So we need to maximize

$$f(a, \ b)=ab.$$

#### Putting the two facts together

The problem with this, is that it is difficult to maximize multivariable functions, and in fact, this function wouldn’t have a maximum unless we go back to our other equation.  Since we know $$a+b=23$$, we can actually use this to rewrite the function we need to maximize in terms of only one variable.

$$a+b=23$$

$$a=23-b$$

Now since we know $$a=23-b$$, we can go to our function of a and b and replace each a with (23-b).  Doing this will give us a new function with only b‘s in it, which we can then maximize.

$$f(a, \ b)=ab$$

Replacing a with (23-b) gives us a new function which we’ll call g.

$$g(b)=(23-b)b$$

#### Maximizing the function to find a and b

Now we just need to find the value of b that maximizes g.  To do this, we just need to take its derivative with respect to b, set it equal to 0, and solve for bThis will always be the general process when you need to maximize or minimize a function.  But first I will expand the function so it’s easier to take its derivative.

$$g(b)=23b-b^2$$

$$g'(b)=23-2b$$

Now set this equal to 0 and solve for b.

$$0=23-2b$$

$$2b=23$$

$$b=\frac{23}{2}$$

We can even check this using Wolfram Alpha.  Below you can see a graph of $$y=g(b)$$ showing the maximum value at $$b=\frac{23}{2}$$.

Now remember, we needed to find the value of a and b that would maximize their product.  So we need to use $$b=\frac{23}{2}$$ to find a.  To do this, we can just use the relationship we found earlier:

$$a=23-b$$

$$a=23-\bigg(\frac{23}{2}\bigg)$$

$$a=\frac{46}{2}-\frac{23}{2}$$

$$a=\frac{23}{2}$$

So now we know that $$a=\frac{23}{2}$$ and $$b=\frac{23}{2}$$ are the two numbers whose sum is 23 and whose product is as large as possible.

There are plenty of other lessons and solutions to help you make sense of derivatives on my derivatives page.  Go check them out and as always, I’d love to here your questions!  Leave a comment below or email me at jakesmathlessons@gmail.com with any questions or suggestions you may have.  Every email and comment helps me gear this site toward what you want to see, so please don’t hesitate to reach out.

You can also enter your name and email below and I’ll send you my calcu 1 study guide full of useful tips and tricks to help you boost your test scores!

## RELATED RATES – 4 Simple Steps

Related rates problems are one of the most common types of problems that are built around implicit differentiation and derivatives.  Typically when you’re dealing with a related rates problem, it will be a word problem describing some real world situation.

Typically related rates problems will follow a similar pattern.  They can usually be broken down into the following four related rates steps:

• The first thing you will usually want to do after reading the problem is to draw a sketch of the situation being described.
• Then you will need to come up with some equation that relates the different quantities described to you, which may be volumes, areas, or distances.
• Once you have this equation, you’ll perform implicit differentiation on both sides of the equation, usually with respect to time.
• Then you just need to solve for the desired rate of change that the question is asking about.

I always think the best way to learn a new concept is practice, practice, practice.  So let’s jump into an example.  If you want to skip ahead, there is a list of other examples at the bottom of this page with a link to their solutions.

## Example 1

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?

### 1. Draw a sketch

Like I said before, the best thing to do first is draw a picture of what is being described.  The problem tells us where these ships are, and the direction and speed at which they’re moving at noon, so I think we should start by drawing our scene at noon, seen in Figure 2.1.

Now let’s think about what will be happening between noon and 4:00 PM.  We know the speed and direction both of these ships are traveling.  We also know that they will be moving for 4 hours before we consider their position again.

Ship A:  This ship is moving 35 km/h for 4 hours.  Therefore, between noon and 4:00 PM we know it will move east

$$35\frac{km}{h}*4h=140km.$$

Ship B:  This ship is moving 25 km/h for 4 hours.  Therefore, between noon and 4:00 PM we know it will move north

$$25\frac{km}{h}*4h=100km.$$

Considering both of these facts, at 4:00 PM, our scene would look something like this:

### 2. Come up with your equation

Now, the next thing we need to do is come up with an equation that relates the different distances given based on where the boats are at 4:00 PM.  We also want to consider what the question is asking before we come up with this equation.  The question is asking us to find how fast the distance between the boats is changing.  We won’t be able to find an equation for this right away, but this tells us that we need an equation that involves the distance between the two boats.  From there, we can figure out how fast that distance is changing.

To do this, we will want to think of the drawing in Figure 2.2 as a triangle.  The three vertices of the triangle would be ship A, ship B, and the point in the water where boat B was at noon, which is shown in the drawing above.  Doing this gives us something like this:

It is important to remember that both ships are still moving, they don’t stop at 4:00.  As a result of this, the sides of our triangle are not constants.  Instead, they would be variables, so the triangle we may want to consider is this one:

From this, we can create our equation.  What we want to think about is what variable has to do with the value the question is asking us to find and what variables do we know something about.

What are we looking for?

The question asks us to find how fast the distance between the ships is changing.  By comparing Figure 2.4 to the prior drawings, we can see that side $$z$$ is the one that represents the distance between the two ships.  Therefore, we will need to include $$z$$ in our equation which we will eventually differentiate.

This is where we want to consider the actual numerical values we have figured out and how they relate to the different variables in our drawing.  At 4:00 PM we know the values of $$x$$ and $$y$$.  We also know the speed at which the ships are moving in relation to a fixed point, which is the vertex of the triangle that makes up the right angle.  Therefore, we not only know the values of $$x$$ and $$y$$, but we also know their rates of change.  These will simply be the speeds of the ships.

Putting it into an equation.

We need an equation that relates the thing we are looking for with the things we already know.  Since we’re looking for some information about $$z$$ (the rate of change of $$z$$), and we know everything about $$x$$, $$y$$, and both of their rates of change at 4:00, we need an equation relating $$x$$, $$y$$, and $$z$$.

Since we know the relationship between these variables is that they are the side lengths of a right triangle, the simplest equation we can use is Pythagorean Theorem.  Based on this we know that

$$z^2=x^2+y^2.$$

### 3. Implicit differentiation

Now that we have our equation, we need to take its derivative.  This is where the implicit differentiation comes in.  Before we do this, let’s think about what we want to differentiate with respect to.

I mentioned in the beginning of this article that we will usually differentiate with respect to time.  The reason for this is that $$x$$, $$y$$, and $$z$$ are all changing as time passes.  If fact, since we know the ships’ initial positions and their velocities, we could actually write $$x$$, $$y$$, and $$z$$ as functions of time.  Therefore, when we differentiate both sides of our equation, we will treat $$x$$, $$y$$, and $$z$$ as functions, and time (represented by $$t$$) will be the variable.  If you want a bit more explanation on the next few step, I explained this a bit more here.

$$z^2=x^2+y^2$$

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

$$2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$$

Now we can divide both sides by $$2$$ to simplify.

$$z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$$

### 4. Solve for desired rate of change

Now remember, the thing we are trying to find in this problem is the rate of change of $$z$$.  This is exactly what $$\frac{dz}{dt}$$ represents, so we will solve for this by dividing both sides by $$z$$.

$$\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}$$

All we have left to do now is plug in all the pieces on the right side of the equation and that would give us our answer.  We are looking for the value of $$\frac{dz}{dt}$$ at 4:00 PM, so we need to use the values of all the other variables based on what they are at 4:00 PM also.  We can gather most of the information we need from Figure 2.2, shown above.  Here it is again:

Clearly, we can see that $$x=10km$$ and $$y=100km$$ at 4:00.  We also know the value of $$\frac{dx}{dt}$$ and $$\frac{dy}{dt}$$. The reason for this is that these two things represent the rate of change of $$x$$ and $$y$$.  Since $$x$$ and $$y$$ represent the distance between each of the ships and a fixed point, $$\frac{dx}{dt}$$ and $$\frac{dy}{dt}$$ would be given by the speed of each ship.  It is also important to point out that is works because the ships are moving directly toward the fixed point or directly away from the fixed point.  The length of $$x$$ is shrinking by $$35\frac{km}{h}$$ because ship A is moving at that speed.  Therefore, we know

$$\frac{dx}{dt}=-35\frac{km}{h}.$$

Notice this value is negative.  This is simply because $$x$$ is getting smaller at 4:00.  Similarly, we know that $$y$$ is getting larger by $$25\frac{km}{h}$$ at 4:00 and therefore we can say that

$$\frac{dy}{dt}=25\frac{km}{h}.$$

Now the only thing we still need to figure out is the value of $$z$$ at 4:00 PM.  To do this, let’s go back to our Pythagorean Theorem equation we looked at earlier.  We know

$$z^2=x^2+y^2.$$

Since we already know the values of $$x$$ and $$y$$ at 4:00, we can just plug them into the Pythagorean Theorem equation to get the value of $$z$$ at 4:00.

$$z^2=10^2+100^2$$

$$z^2=100+10,000$$

$$z^2=10,100$$

$$z=\sqrt{10,100}$$

Now that we have all the pieces figured out, we can go back and plug them into our equation for $$\frac{dz}{dt}$$.

$$\frac{dz}{dt}=\frac{10*(-35)+100*25}{\sqrt{10,100}}$$

$$\frac{dz}{dt}=\frac{2,150}{\sqrt{10,100}}\approx 21.393\frac{km}{h}$$

So we can say that the distance between the two ships is changing at about $$21.393\frac{km}{h}$$ at 4:00 PM.

## Other Examples

Like I said before, the best way to gain an understanding of related rates problems is practice.  Here are some more complete solutions of other fun related rates problems.  Just click on the problem to see the full solution.

#### Triangles

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?

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?

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?

#### Squares

Each side of a square is increasing at a rate of 6 cm/s. At what rate is the area of the square increasing when the area of the square is 16 cm^2?

#### Cones

Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time water is being pumped into the tank at a constant rate. The tank has a height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

#### Spheres

If a snowball melts so that its surface area decreases at a rate of 1 cm^2/min, find the rate at which the diameter decreases when the diameter is 10 cm.

The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 80 mm?

#### Cylinders

A cylindrical tank with radius 5 m is being ﬁlled with water at a rate of 3 m^3/min. How fast is the height of the water increasing?

As always, I’d love to see your questions!  You can leave a comment below or email your questions to me at jakesmathlessons@gmail.com.  Whether you have questions about this lesson or want me to post a solution for another problem you’re stuck on, just send me a message.

## Implicit Differentiation

Before getting into implicit differentiation, I would like to take some time to discuss variables, functions, and constants.  The reason for this is that when you do an implicit differentiation problem, you will likely be dealing with equations containing multiple letters.

Up to this point, most of the functions you have taken the derivative of usually contain one variable, usually $$x$$, and any other letters in the function would be constants.  But with implicit differentiation, you will also need to deal with having another letter that represents another unknown function.

Before you start implicitly differentiating a problem I recommend determining whether each letter represents a function, or if it’s a variable or a constant.  This is because each one will be treated differently when you take its derivative.  Since implicit differentiation is essentially just taking the derivative of an equation that contains functions, variables, and sometimes constants, it is important to know which letters are functions, variables, and constants, so you can take their derivative properly.

In many cases, the problem will tell you if a letter represents a constant.  If a letter is a constant, that means you would treat it like it’s a number.  If this is a little confusing, just imagine what would happen if you were to actually put some number in for the constant and think about what would happen to that number when you take the function’s derivative.  Then revert back to having the letter in the equation and treat the constant the same way you treated the number.

Let’s jump into an example and I will explain the process along the way.

## Example 1

Find the derivative of $$f(x)=cx^2+d$$ where $$c$$ and $$d$$ are constants.

#### What to do with constants?

Like I said before, since $$c$$ and $$d$$ are constants, we can treat them as if they are just some number and take the derivative of the remaining function with $$x$$ being the variable.

Let’s imagine $$c$$ and $$d$$ have been replaced with $$2$$ and $$4$$ respectively, and see what happens.

$$f(x)=2x^2+4$$

This is a case where we can just use the power rule to find:

$$f'(x)=2(2x)=4x$$

Now, we can revert back to having $$c$$ and $$d$$ in our function and we would see that:

$$f(x)=cx^2+d$$

$$f'(x)=c(2x)$$

$$f'(x)=2cx$$

Notice, the $$d$$ disappeared because the derivative of a constant is just $$0$$.

#### What about functions and variables?

Now that we have discussed some methods for identifying constants and how to deal with them when taking a derivative, I will discuss indicators for classifying letters that represent functions and those that are variables.

Frequently, when doing an implicit differentiation problem, you will simply be asked to find $$\frac{dy}{dx}$$.  Then you will be shown some equation that contains at least one $$y$$ and at least one $$x$$.  Although this seems like you haven’t been given much direction, the $$\frac{dy}{dx}$$ is actually an indicator that gives us all the information we need.  This notation is one way to write:

The derivative of $$y$$ with respect to $$x$$.

Or in other words, it’s telling you that you are trying to find the derivative of your function, $$y$$, with respect to the variable, $$x$$.  Which tells you that $$y$$ is a function of $$x$$.

In fact, this notation will always give you those two pieces of information.  For example, $$\frac{dh}{dt}$$ is a symbol that represents “the derivative of $$h$$ with respect to $$t$$.”  Therefore, $$h$$ must be a function and $$t$$ must be its variable.

## Example 2

Find $$\frac{dy}{dx}$$ if $$y^2=4x^5-e^x$$.

In a problem like this, since we know we need to find $$\frac{dy}{dx}$$ and we are given an equation which relates $$y$$ and $$x$$, we need to find the derivative of both sides of the equation.

$$y^2=4x^5-e^x$$

Now we can take the derivative of both sides.  Remember, as long as we do the same thing to both sides of an equation, the results will be equal to each other also.

$$\frac{d}{dx}\big[y^2\big]=\frac{d}{dx}\big[4x^5-e^x\big]$$

#### The Left Side of the Equation:

The $$\frac{d}{dx}$$ just means that you need to take the derivative of whatever follows, treating $$x$$ as the variable.  The left side of this equation is the tricky part.  Since the question told us to find $$\frac{dy}{dx}$$, we know that $$y$$ is a function of $$x$$.  The fact that $$y$$ is a function tells us that we can’t just use the power rule to find the derivative of the left side of the equation.  We will actually need to use the chain rule.

We need to do the chain rule because $$y$$ is not a variable here.  Since $$y$$ is a function of $$x$$ and we are taking the derivative with respect to $$x$$, we cannot say that the derivative of $$y^2$$ is $$2y$$!  Now that we have determined that we need to use the chain rule, we need to determine our inside and outside functions.  Remember, we need to figure out some $$f(x)$$ and a $$g(x)$$ so that $$f\big(g(x)\big)=y^2$$ (if you need a refresher on the chain rule, click here).

Typically, when we have a letter that represents a function and we take its derivative with respect to a different variable, we can call our inside function just the single letter which represents a function.  Therefore, we can say our inside function is $$g(x)=y$$.

Now, to find our outside function, we can look at the entire function and replace the inside function with a single $$x$$.  We are replacing it with an $$x$$ because that is the variable we are differentiating with respect to.  So if we take our function ($$y^2$$) and replace the inside function ($$y$$) with a single $$x$$, we are left with our outside function $$f(x)=x^2$$.  At this point we have figured out:

$$f(x)=x^2,$$

$$g(x)=y.$$

The next thing we need to do is find the derivative of both our inside and outside functions.  Finding $$f'(x)$$ can be found simply using the power rule:

$$f'(x)=2x.$$

Now we need to find $$g'(x)$$.  This is a little more tricky.  The key thing, which I will continue to remind you of, is that we are taking the derivative of $$y$$ with respect to $$x$$.  Therefore, we cannot say that the derivative of $$y$$ is $$1$$.

In fact, we do not know the derivative of $$y$$.  Since $$y$$ is some function of $$x$$ that we actually don’t know, we can’t explicitly write its derivative either.  But luckily, we don’t need to be able to do this.  All we need to say is that the derivative of $$y$$ is the symbol I mentioned earlier which represents “the derivative of $$y$$ with respect to $$x$$.”  We can simply use $$\frac{dy}{dx}$$ to represent this.  Therefore, we know that:

$$g'(x)=\frac{dy}{dx}.$$

Now we can just use these pieces and plug them into the chain rule formula.

$$\frac{d}{dx}\big[y^2\big]=f’\big(g(x)\big)\cdot g'(x)$$

$$\frac{d}{dx}\big[y^2\big]=2(y)\cdot \frac{dy}{dx}$$

$$\frac{d}{dx}\big[y^2\big]=2y\frac{dy}{dx}$$

#### The Right Side of the Equation:

Finding $$\frac{d}{dx}\big[4x^5-e^x\big]$$ is quite a bit easier than the left side of the equation.  Since this side contains no other letters besides $$x$$, which is the variable we are differentiating with respect to, this will be like any other derivative we have taken up to this point.

$$\frac{d}{dx}\big[4x^5-e^x\big]=4(5x^4)-e^x$$

$$\frac{d}{dx}\big[4x^5-e^x\big]=20x^4-e^x$$

#### Putting It All Together:

Back to our original equation, we had:

$$\frac{d}{dx}\big[y^2\big]=\frac{d}{dx}\big[4x^5-e^x\big].$$

And as we just showed above, this means:

$$2y\frac{dy}{dx}=20x^4-e^x.$$

Now, once we have taken the derivative of both sides, you can see that our equation contains a $$\frac{dy}{dx}$$.  Since our goal here is to find $$\frac{dy}{dx}$$, now that we have an equation that contains it, all we have to do is solve for $$\frac{dy}{dx}$$.

All we have to do is divide both sides by $$2y$$.

$$\frac{2y\frac{dy}{dx}}{2y}=\frac{20x^4-e^x}{2y}$$

Once we simplify the left side we are left with

$$\frac{dy}{dx}=\frac{20x^4-e^x}{2y}.$$

Although this looks a little strange, since our equation for $$\frac{dy}{dx}$$ contains both $$x$$ and $$y$$, this is sometimes the best we can do.  Implicit differentiation is most useful in the cases where we can’t get an explicit equation for $$y$$, making it difficult or impossible to get an explicit equation for $$\frac{dy}{dx}$$ that only contains $$x$$.  Therefore, we have our answer!

I would like to point out that this example is actually a case where we could have solved for $$y$$ in terms of $$x$$ before taking the derivative.  Doing this would have meant that we could have used other derivative tricks and avoided implicit differentiation, but the way I solved it shows the process of implicit differentiation which is applicable in cases where it is absolutely necessary.

In those cases the general idea and process is the same: we have some function that relates $$y$$ and $$x$$ and we need to take the derivative of both sides, then use algebra to solve for $$\frac{dy}{dx}$$.  This may not always be as simple as the above example, but the process will be extremely similar.

## Example 3

Find $$\frac{dy}{dx}$$ if $$y=x^x$$.

This problem is going to be a bit more tricky than the first two examples.  Click here to see the full solution.

## More Examples

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

As always, don’t forget to let me know if you have any questions on this lesson or if you have any suggestions for other lessons you want to see in the future.  Go check out my derivatives page to see what other material I’ve covered.  I want to know what you want to see on this site, so any and all suggestions and questions are welcome if you can’t find an answer to your question in another lesson.  Just go ahead and leave a comment on this post or email me at jakesmathlessons@gmail.com!