implicit Archives | Jake's Math Lessons

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!

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?

Jake’s Math Lessons Complete Calculus 1 Package

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.

A kite 100ft above the ground moves horizontally at a speed of 8 ft/s.

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.

What do we know about?

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:

Angle $\theta$ (this will be changing as the kite moves).
Side $a$ (this will be changing as the kite moves and the string is let out).
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
Cosine Adjacent Hypotenuse
Tangent Opposite Adjacent

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.

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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.

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?