Introduction- Implicit Differentiation

[EDITOR'S NOTE: For clarity, we will be using Leibniz Notation to denote the derivative of $y$ with respect to $x$ $\left(\frac{dy}{dx}\right)$ for Implicit Differentiation. Note that the Lagrangian style $(y^{\prime })$ can also be used as well, but if written poorly on paper, can cause confusion with the $y$ terms proper. To avoid this issue entirely, we will be using Leibniz.]

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

$y=x^2$

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

$y=\sqrt{x-2}$

As you can see all of these functions have an independent variable that's $x$ and a dependent variable 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:

$x-y^2=1$

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

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

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(s) 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. The idea of implicit differentiation is that you can differentiate an implicit equation. An implicit equation is something that isn’t in the form of $y=f(x)$.  You can still find the derivative of these functions using a special application of the chain rule called implicit differentiation.

Intuition Behind Implicit Differentiation

The idea of implicit differentiation is that it takes an implicit equation, differentiates it, and finds the derivative. The idea is based on the chain rule.

Let’s have two functions, $f(x)=x$ and $g(y)=y$. The idea of this is so that we have $f(g(y))=y$. If we apply the chain rule, we get $f^{\prime }(g(y))\cdot g^{\prime }(y)=f^{\prime }(y)\cdot g^{\prime }(y)$. If we take the derivative of $f(x)=x$ and $g(y)=y$, we get $f^{\prime }(x)=1$ and $g^{\prime }(y)=\frac{dy}{dx}$. 

Why does differentiating $g(y)=y$ lead to $g^{\prime }(y)=\frac{dy}{dx}$ and not $g^{\prime }(y)=1$ like how $f^{\prime }(x)$ is? Well, recall that whenever you differentiate $f(x)$, it turns into $f^{\prime }(x)$. Using $y$ and $y^{\prime }$ is different notation that represents the same thing. 

Now to figure out $f^{\prime }(y)$, if $f^{\prime }(x)=1$, then $f^{\prime }(y)=1$ too.

If we plug everything in, we get $f^{\prime }(y)\cdot \frac{dy}{dx}=1\cdot \frac{dy}{dx}=\frac{dy}{dx}$

$\frac{dy}{dx}$ is the result of differentiating $y$. This may all seem a bit confusing, but let’s change up $f(x)$.

If we had two functions, $f(x)=x^3$ and $g(y)=y$, we can set up that $f(g(y))=y^3$ and differentiate using the chain rule.

$f(x)=x^3$

$g(y)=y$

$f^{\prime }(x)=3x^2$

$g^{\prime }(y)=\frac{dy}{dx}$

Plugging all of this into the chain rule gives us the answer.

$f^{\prime }(g(y))\cdot g^{\prime }(y)=f^{\prime }(y)\cdot \frac{dy}{dx}$

If $f^{\prime }(x)=3x^2$, then $f^{\prime }(y)=3y^2$. 

$f^{\prime }(y)\cdot \frac{dy}{dx}=3y^2\cdot \frac{dy}{dx}$

The pattern is that when you differentiate anything with a $y$ variable, you multiply that term by $\frac{dy}{dx}$.

Examples of Implicit Differentiation

Let’s get the gist of this by implicitly differentiating some stuff.

 Example #1

Find the derivative of $x^2+y^2=1$.

We can differentiate everything normally, like this!

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

Notice how when we differentiated the $y$ term, we multiplied that term by $y’$. The only problem is, we haven’t found the derivative. The derivative is equal to $y’$, and so we need to solve for $\frac{dy}{dx}$.

$2x+2y\cdot \frac{dy}{dx}=0⟹2y\cdot \frac{dy}{dx}=-2x⟹\frac{dy}{dx}=\frac{-2x}{2y}⟹\frac{dy}{dx}=\frac{-x}{y}$

Let’s do another one!

Example #2

Find the tangent line at $(0,-2.64575)$ for the implicit equation $x^{2}y+y^2=7$.

For this one, we can differentiate as normal, nothing complicated. However, as we discussed before, you must multiply by $\frac{dy}{dx}$ whenever you are differentiating.

For the first term $x^{2}y$, there is a product between two functions, so let us do the product rule.

$f(x)=x^2$

$g(y)=y$

$f^{\prime }(x)=2x$

$g^{\prime }(y)=\frac{dy}{dx}$

Plug this into the well familiarized product rule, I don’t think this needs to be repeated.

$(2x)(y)+(x^2)\left(\frac{dy}{dx}\right)$

Remember that this is only one term, let’s now differentiate $y^2$, which turns out to be $2y\cdot \frac{dy}{dx}$. See that you’re differentiating the same exact way, just with a different variable and a slightly different rule. 

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

So this is our answer! NOT! Again, we need to find $\frac{dy}{dx}$, not just differentiate everything. Let’s just commit the act of simple algebra.. Wait. Hold up… didn’t we start off with an equation? OH, we forgot the $7$. Let’s differentiate it and get that the answer is 0.

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

Finally, we are finished differentiating! Let’s solve for the derivative now.

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

Nice! WAIT. NOT NICE. We are asked to find the tangent line at a point! Well my friend, we have the derivative, and we have the numbers we need, so let’s just find the slope as usual!

Notice how we are given the $x$ and $y$ coordinates, and so we can simply plug them into our derivative to find the slope of the tangent line at $(0,-2.64575)$.

$\frac{dy}{dx}=\frac{2(0)(-2.64575)}{(0{)}^2+2(-2.64575)}=0$

Would you take a look at that? The slope is $0$. Well, let’s just plug that into our point slope formula and solve!

$y=m(x-x_1)+y_1⟹y=0(x-0)+(-2.64575)⟹y=-2.64575$

This is our tangent line, nice!

Monstrous Example

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

Try differentiating this yourself again before looking over my steps.

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

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

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

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

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

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

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

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

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

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

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

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

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. For the sake of everyone's sanity, we'll do all the further math necessary and present it in a more cleaner, simplified form. Here it is:

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

Some Notes On Implicit Differentiation

Practice Questions