Introduction - What is Implicit Differentiation?

Up to this point you have only differentiated functions with respect to their input variable; these functions are explicit functions. Some examples of explicit functions include the following:

$y=x^2$

$f(x)=\sin (\ln (x))$

$x=\sqrt{y-2}$, Editor's Note: This is an explicit function of $x$ with respect to $y$

As you can see all of these functions have an independent variable for the first example that's $x$ and a dependent variable, for the first example that is $y$. You have learned how to differentiate a variety of explicit functions. What we are going to be covering in this article is how to differentiate implicit functions. Implicit functions can look like these examples of implicit functions:

$x-y^2=1$

$\tan (xy)=\sin (y)-x{y}^5$

$y-\ln (xy)=4x+1$

These, implicit functions, are functions not defined with respect to a single variable. Some of these functions you can rewrite as explicit functions, the first function can be rewritten as an explicit function of $x$:

$x-y^2=1$

$y^2=x-1$

$y=\pm \sqrt{x-1}$

Or as a function of $y$:

$x-y^2=1$

$x=1+y^2$

There are some functions, such as the second example I gave ($\tan (xy)=\sin (y)-x{y}^5$), where it is not possible, or extremely difficult, to express them as explicit functions. You can still find the derivative of these functions using a special application of the chain rule called implicit differentiation.

An Example of Implicit Differentiation

Let's try differentiating that equation. First apply the differential with respect to $x$ to both sides of the equation:

$\frac{d}{dx}(x-y^2)=\frac{d}{dx}(1)$

Then apply the difference rule to the left side of the equation and the constant rule to the right.

$\frac{d}{dx}(x)-\frac{d}{dx}(y^2)=0$

We know the derivative of $x$ with respect to $x$ is just 1.

$1-\frac{d}{dx}(y^2)=0$

Now we have to find the derivative of $y^2$ with respect to $x$. This is where we apply the chain rule. You can think of the inner function here as $y$ and the outer function as the square function.

$u=y$

$v=x^2$

$1-v’(y)\cdot \frac{d}{dx}(y)=0$

$1-2y\cdot \frac{dy}{dx}=0$

Now to find the derivative we have to solve for $\frac{dy}{dx}$.

$1=2y\cdot \frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{2y}$

That's the basic process for finding a derivative using implicit differentiation. The derivative we found for this equation was in terms of $y$ but you could also find a derivative in terms of $x$ and $y$. I’ll quickly go over another example differentiating:

$x^{2}y+y^2=7$

I encourage you to give it a try before looking over my steps.

$x^{2}y+y^2=7$

$\frac{d}{dx}(x^{2}y+y^2)=\frac{d}{dx}(7)$

$\frac{d}{dx}(x^{2}y)+\frac{d}{dx}(y^2)=0$

$(2xy+x^2\frac{dy}{dx})+2y\frac{dy}{dx}=0$

$2xy+x^2\frac{dy}{dx}+2y\frac{dy}{dx}=0$

$x^2\frac{dy}{dx}+2y\frac{dy}{dx}=-2xy$

$(x^2+2y)\cdot \frac{dy}{dx}=-2xy$

$\frac{dy}{dx}=-\frac{2xy}{x^2+2y}$

This example was quite a bit more complicated as it involved the product rule and quite a bit more rearranging to solve but it's the same general process for implicit differentiation as the first example. Let's go through one final difficult example, the equation we will differentiate this time is:

$\sin \left(\frac{x}{y}\right)+\ln (y)=e^{xy}-21$

Try differentiating this yourself again before looking over my steps.

$\sin \left(\frac{x}{y}\right)+\ln (y)=e^{xy}-21$

Try differentiating this yourself again before looking over my steps.

$\sin \left(\frac{x}{y}\right)+\ln (y)=e^{xy}-21$

$\frac{d}{dx}[\sin \left(\frac{x}{y}\right)+\ln (y)]=\frac{d}{dx}[e^{xy}-21]$

$\cos \left(\frac{x}{y}\right)\cdot \frac{d}{dx}(\frac{x}{y})+\frac{1}{y}\cdot \left(\frac{dy}{dx}\right)=e^{xy}\cdot \left(\frac{dy}{dx}\right)(xy)$

$\cos \left(\frac{x}{y}\right)\cdot \frac{y-x\left(\frac{dy}{dx}\right)}{y^2}+\frac{1}{y}\left(\frac{dy}{dx}\right)=e^{xy}\cdot (y+x\left(\frac{dy}{dx}\right))$

$\cos \left(\frac{x}{y}\right)\cdot \frac{y-x\left(\frac{dy}{dx}\right)}{y^2}+\frac{1}{y}\left(\frac{dy}{dx}\right)=y{e}^{xy}+x{e}^{xy}\left(\frac{dy}{dx}\right)$

$\cos \left(\frac{x}{y}\right)\cdot \frac{y-x\left(\frac{dy}{dx}\right)}{y^2}=y{e}^{xy}+x{e}^{xy}\left(\frac{dy}{dx}\right)-\frac{1}{y}\left(\frac{dy}{dx}\right)$

$\frac{\cos \left(\frac{x}{y}\right)}{y^2}(y-x\left(\frac{dy}{dx}\right))=y{e}^{xy}+x{e}^{xy}\left(\frac{dy}{dx}\right)-\frac{1}{y}\left(\frac{dy}{dx}\right)$

$y\frac{\cos \left(\frac{x}{y}\right)}{y^2}-x\frac{\cos (\frac{x}{y})}{y^2}\left(\frac{dy}{dx}\right)=y{e}^{xy}+x{e}^{xy}\left(\frac{dy}{dx}\right)-\frac{1}{y}\left(\frac{dy}{dx}\right)$

$y\frac{\cos \left(\frac{x}{y}\right)}{y^2}-y{e}^{xy}=x\frac{\cos (\frac{x}{y})}{y^2}\left(\frac{dy}{dx}\right)+x{e}^{xy}\left(\frac{dy}{dx}\right)-\frac{1}{y}\left(\frac{dy}{dx}\right)$

$y\frac{\cos (\frac{x}{y})}{y^2}-y{e}^{xy}=\left(\frac{dy}{dx}\right)\cdot (x\frac{\cos (\frac{x}{y})}{y^2}+x{e}^{xy}-\frac{1}{y})$

$\frac{y\frac{\cos (\frac{x}{y})}{y^2}-y{e}^{xy}}{x\frac{\cos (\frac{x}{y})}{y^2}+x{e}^{xy}-\frac{1}{y}}=\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{y\frac{\cos \left(\frac{x}{y}\right)}{y^2}-y{e}^{xy}}{x\frac{\cos \left(\frac{x}{y}\right)}{y^2}+x{e}^{xy}-\frac{1}{y}}$

That example was really difficult but it’s the same process that we used before. Now let's go over some final notes before we get to the practice questions.

Some Notes On Implicit Differentiation

Practice Questions