• 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.

Assignment

  • Vocabulary and teal boxes
  • p180 1–21 odd, 25, 28–31, 42, 43, 47, 49, 50, 53, 55, 59–61, 63, 68 80, 81, 84–87

Additional Resources

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

  1. Differentiate both sides with respect to $x$.
  2. Collect all terms involving $\frac{dy}{dx}$ on one side, with the rest going on the other.
  3. Factor out $\frac{dy}{dx}$.
  4. 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.