Kurzius Math Notes

2.5 Implicit Differentiation

AP Topics: 3.2, 5.12

Distinguish between functions written in implicit form and explicit form.
Use implicit differentiation to find the derivative of a function.
Find derivatives of functions using logarithmic differentiation.

✎ p180 1–21 odd, 25, 28–31, 42, 43, 47, 49, 50, 53, 55, 59–61, 63, 68 | 80, 81, 84–87

Up until this point, we’ve differentiated explicit functions, which is when a function $y$ is written as a function of $x$. Like, $y=1/x$. An implicit function is when this isn’t the case. If we rearranged that equation, we can get the implicit function $xy=1$.

So, we’ll be looking at how to differentiate functions that can’t be defined explicitly. Equations like $y^2=x$.

Before that, let’s revisit chain rule and the power rule. And let’s be really picky and define our inner function as $u=x$.

\[\begin{align*} \frac{d}{dx}[x^2] &= 2x\frac{d}{dx}[x] \\[1em] &= 2x\frac{dx}{dx} \\[1em] &= 2x \end{align*}\]

Since we are always differentiating with respect to $x$, those steps are unnecessary. There’s no need to explore how $x$ changes with respect to $x$.

We’ll take this same idea over to implicit functions. The variable itself will be a function that needs to be differentiated with respect to $x$.

Let’s differentiate $y^2=x$.

\[\begin{align*} \frac{d}{dx}[y^2]&=\frac{d}{dx}[x] \\[1em] 2y\frac{d}{dx}[y] &= 1 \\[1em] 2y\frac{dy}{dx} &= 1 \\[1em] \frac{dy}{dx} &= \frac{1}{2y} \end{align*}\]

Note that last step has us explicitly define the derivative by isolating $\frac{dy}{dx}$.

There’s a general strategy on page 175 in the book that I’ll include here.

Guidelines for Implicit Differentiation

Differentiate both sides with respect to $x$.

Collect all terms involving $\frac{dy}{dx}$ on one side, with the rest going on the other.

Factor out $\frac{dy}{dx}$.

Solve for $\frac{dy}{dx}$.

When finding higher order derivatives, you can substitute in early ones when needed. Let’s find the second derivative of $x^2+y^2 = 25$.

\[\begin{align*} \frac{d}{dx}[x^2 + y^2] &= \frac{d}{dx}[25] \\[1em] 2x +2y\frac{dy}{dx} &= 0 \\[1em] 2y\frac{dy}{dx} &= -2x \\[1em] \frac{dy}{dx} &= \frac{-2x}{2y} \\[1em] \frac{dy}{dx} &= -\frac{x}{y} \\[1em] \end{align*}\]

Now for number 2.

\[\begin{align*} \frac{d^2y}{dx^2}&=-\frac{1\cdot y-x\frac{dy}{dx}}{y^2} \\[1em] &= \frac{y-x\left(-\frac{x}{y}\right)}{y^2} \\[1em] &= \frac{y+\frac{x^2}{y}}{y^2} \\[1em] &= \frac{\frac{y^2}{y}+\frac{x^2}{y}}{y^2}\\[1em] &= \frac{y^2+x^2}{y^3} \end{align*}\]

Quick note: you will not be able to graph derivatives of implicit functions in most cases.

Logarithmic Differentiation

This is a strategy for differentiating functions that would otherwise involve a heavy use of product or quotient rule, or when you have a variable in both the base and exponent, like $x^x$. It makes use of these log rules.

\[\ln\left(ab\right) = \ln a + \ln b \qquad \ln\left(\frac{a}{b}\right) = \ln a - \ln b \qquad \ln a^b = b\ln a\]

We’ll cover some examples in class, but generally speaking, take the natural log of both sides, use the appropriate log rule to rewrite, then differentiate as usual.

Using this is a personal preference and it is not directly tested on the AP exam.