Implicit Differentiation | Jake's Math Lessons

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!