Implicit Differentiation

Until this point, we’ve been concerned with derivatives of functions like $y=x$ or $f(x)=x$; essentially, functions that are defined with the dependent variable isolated on one side, with only one variable on the other side. We’ve been taking derivatives and calling them for the most part things like $y^{\prime }$ (”$y$ prime”) or $f^{\prime }(x)$ (”$f(x)$ prime”), but that’s just because we’ve restricted ourselves to Lagrange’s notation for derivatives.

However, for functions and non-functions that aren’t explicit like the previous examples, we’ll have to dive a little deeper into Leibniz’s notation and implicit differentiation.

Leibniz’s notation review §

We’ve elaborated on this about a quarter of a billion times, but I’ll say it again; a derivative is an equation thats output is the rate of change or slope of the function it is a derivative of— its primitive, its antiderivative, there’s a lot of names for it.

Going back to our elementary school knowledge of slope, we know that we can find the slope $m$ using the following equations:

$$m=\frac{\text{rise}}{\text{run}}\mid m=\frac{\Delta y}{\Delta x}$$

Here the capital delta ($\Delta$) means “change in;” slope is equal to the change in $y$ over the change in $x$.

Leibniz’s notation is a literal implementation of this. If I were to represent the derivative of $y=x^2$ in Leibniz’s notation, I would write it $\frac{dy}{dx}$; the derivative of $y$ with respect to $x$, or the difference in $y$ over the difference in $x$. So, $y=x^2$, $\frac{dy}{dx}=2x$. Leibniz’s notion is particular useful in that it helps us remember that a derivative is still essentially that same as the concept of a slope that we know very well— we just have a new method of finding it.

What and how? §

Let’s consider an equation for an ellipse:

$$\frac{(x-2{)}^2}{4}+\frac{(y+1{)}^2}{9}=1$$

This equation is not explicitly set up for you to find some $y$ or $f(x)$ or something explicitly. While you could theoretically find the $y$ coordinates that match with $x$ coordinates, the fact that we raise $x$ and $y$ to the second power makes it impossible to find only one coordinate; you’ll get two.

This is where implicit differentiation comes in. Since we need both an $x$ and $y$ coordinate to figure out where we are now, we can’t just find the rate of change at a specific $x$ coordinate and call it a day; we need to have an equation that also takes a $y$ coordinate in consideration.

Differentiating §

Consider the equation of an ellipse again.

$$\frac{(x-2{)}^2}{4}+\frac{(y+1{)}^2}{9}=1$$

Implicit differentiation is quite similar to regular derivation in that we follow the same rules for individual terms and expressions. To find the derivative of $y$ with respect to $y$, we apply these rules with respect to the derivative of $\frac{d}{dx}$:

$$\begin{array}{rl}\frac{d}{dx}\left(\frac{(x-2{)}^2}{4}\right)+\frac{d}{dx}\left(\frac{(y+1{)}^2}{9}\right)& =\frac{d}{dx}(1)\\ \frac{1}{4}\cdot 2(x-2)+\frac{1}{9}\cdot 2(y+1)\cdot \frac{dy}{dx}& =0\\ \frac{1}{2}(x-2)+\frac{2}{9}(y+1)\cdot \frac{dy}{dx}& =0\end{array}$$

Solving for $\frac{dy}{dx}$:

$$\begin{array}{rl}\frac{2}{9}(y+1)\cdot \frac{dy}{dx}& =-\frac{1}{2}(x-2)\\ \frac{dy}{dx}& =-\frac{9}{4}\cdot \frac{x-2}{y+1}\\ & =\frac{-9x-18}{4(y+1)}\end{array}$$

Now, $\frac{dy}{dx}$ is expressed in terms of both $x$ and $y$, providing the slope of the tangent line to the ellipse at any given point $(x,y)$.

Higher Implicit Differentiation §

Sometimes we need more than just the first derivative of a function, though. Let’s say we want to go from the first derivative $\frac{dy}{dx}$ above and find the second derivative $\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d^{2}y}{d{x}^2}$, or $\frac{d}{dx}\left[-\frac{9}{4}\cdot \frac{x-2}{y+1}\right]$. We can use a similar approach here as we did above.

$$\begin{array}{rl}\frac{d}{dx}\left(\frac{dy}{dx}\right)& =\frac{d}{dx}\left[\frac{-9x-18}{4(y+1)}\right]\\ & \\ & =\frac{\frac{d}{dx}[-9x+18][4(y+1)]-\frac{d}{dx}[4(y+1)](-9x-18)}{4(y+1{)}^2}\\ & =\frac{-9(1+y)-\frac{dy}{dx}(-9x-18)}{4(y+1{)}^2}\end{array}$$

You might realize now that we’ve managed to create $\frac{dy}{dx}$ out of one of our $\frac{d}{dx}$ terms, which means we can substitute our known value $\frac{dy}{dx}$ in.

$$\begin{array}{rl}\frac{d}{dx}\left(\frac{dy}{dx}\right)& =\frac{-9(1+y)-\left(\frac{-9x-18}{4(y+1)}\right)(-9x+18)}{4(y+1{)}^2}\\ & =\frac{-81x^2-324x-36y^2-72y-360}{16(1+y{)}^3}\\ & =-\frac{9(9x^2+36x+4y^2+8y+40)}{16(1+y{)}^3}\end{array}$$

That kind of sucks though.

How else can we use implicit differentiation? §

Implicit differentiation is a cool tool that we can use not just to express derivatives of equations but unknown derivatives. Let’s say we have an equation $u^2-x^2+y^2=0$, and we want to find the rate of change of $u$ with respect to $x$. First, we have to differentiate each term with respect to x.

$$\begin{array}{rl}\frac{d}{dx}(u^2)-\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)& =\frac{d}{dx}(0)\\ 2u\frac{du}{dx}-2x+2y\frac{dy}{dx}& =0\end{array}$$

Now that we have the equation differentiated with respect to $x$, we can find the derivative of $u$ with respect to $x$ by solving for $\frac{du}{dx}$.

$$\begin{array}{rl}2u\frac{du}{dx}& =2x-2y\frac{dy}{dx}\\ \frac{du}{dx}& =\frac{2x-2y\frac{dy}{dx}}{2u}\end{array}$$

If we knew the derivative of $\frac{dy}{dx}$, we could just substitute that in there. and solve through— a similar approach as finding higher derivatives.