• 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 (34 problems)

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 $xy=1$.

But the function above can be written either explicitly or implicitly. We’ll be looking at how to differentiate functions that can’t be defined explicitly. Equations like $y^2=x$.

Before that, there are two things that needed to be point out. First, let’s revisit chain rule and do it while defining our inner function as ${u=x}$.

\[\begin{align} \frac{d}{dx}[u^2] &= 2u\frac{d}{dx}[u] \\[1em] &= 2u\frac{du}{dx} \\[1em] &= 2x\frac{dx}{dx} && \text{Substitute $u=x$} \\[1em] &= 2x \end{align}\]

Since we are always differentiating with respect to $x$, those extra steps are unnecessary. It’s the equivalent of asking how $x$ changes with respect to $x$. Well, at a rate of $1$ for every $1$.

The second thing that needs to be brought up is that differentiating is an operation performed on both sides of an equation.

\[\begin{align} y &= x^2 + 4 \\ \frac{d}{dx}[y] &= \frac{d}{dx}[x^2+4] \\ \frac{dy}{dx} &= 2x \end{align}\]

Again, this was going on behind the scenes, but not worth mentioning before. Sorry about that.

Now with all that out in the open, let’s differentiate an implicit function. We just have to make sure to differentiate both sides and to use the chain rule on everything.

Example 1

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

\[\begin{align} \frac{d}{dx}[y^2]&=\frac{d}{dx}[x] && \text{Differentiate both sides}\\[1em] 2y\frac{d}{dx}[y] &= 1 &&\text{Chain rule applied to $y$} \\[1em] 2y\frac{dy}{dx} &= 1 \\[1em] \frac{dy}{dx} &= \frac{1}{2y} && \text{Solve for $dy/dx$} \end{align}\]

$\blacksquare$

There’s a general strategy for implicit differentiation on page 175 in the book that I’ll include here as well. It will be helpful for the next example.

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 $\frac{dy}{dx}$.

Example 2

Find $dy/dx$ given that $y^3+y^2-5y-x^2=-4$.

\[\begin{align} \frac{d}{dx}[y^3+y^2-5y-x^2]&=\frac{d}{dx}[-4] \\ 3y^2\frac{dy}{dx} + 2y\frac{dy}{dx}-5\frac{dy}{dx}-2x&=0 \\ 3y^2\frac{dy}{dx} + 2y\frac{dy}{dx}-5\frac{dy}{dx}&=2x \\ \frac{dy}{dx}(3y^2 + 2y-5)&=2x \\ \frac{dy}{dx} &= \frac{2x}{3y^2 + 2y-5} \end{align}\]

$\blacksquare$

When finding higher order derivatives, you can substitute in earlier ones as needed.

Example 3

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.

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

$\blacksquare$

Quick note: you won’t be able to graph the derivatives of implicit functions easily. It requires breaking the original implicit functions into multiple explicit functions and then substituting them back into $dy/dx$. The algebra required to this can sometimes be way beyond anything we’ve covered, but there are some simple ones you can do, like the first derivative of example 3.

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 $x^x$. Using this is a personal preference and is not directly tested on the AP exam.

We’ll cover an example or two class, but generally the trick is to find the natural log of both sides, and then use the appropriate log rule to rewrite, then differentiate as usual.