• Find the derivative of an inverse function.
  • Differentiate an inverse trigonometric function.

Assignment

  • Vocabulary and teal boxes
  • p187 1, 5, 8, 10, 11, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40–45, 48, 52, 59, 61 | 76, 85–88

The vertical bar indicates the AP cut-off. Problems after the bar are recommended for AP prep, but not required for the assignment.

Additional Resources

Inverse functions are when your $x$ and $y$ values get reversed (lesson from chapter P here). Differentiating an inverse function isn’t any different than any other function, but there is a relationship between the two derivatives so that one can tell you about the other. We’ll use property of inverse functions were composing them returns your original $x$.

\[\begin{align*} f(g(x)) &= z \\ \frac{d}{dx}[f(g(x))] &= \frac{d}{dx}[x]\\ f'(g(x)) \cdot g'(x) &= 1 \\ g'(x) &= \frac{1}{f'(g(x))} \end{align*}\]

This relationship makes it easy to find the derivative of the inverse at a point, even when the function is unknown. And in fact, most problems involving derivatives and inverses provide points for you. There is no need to figure out the inverse function and its derivative.

For example, let $f(x) = \frac{1}{4}x^3 + x - 1$ and we want to find $(f^{-1})’(3)$. Start with the equation and then determine what information you need to find.

\[(f^{-1})'(3) = \frac{1}{f'(f^{-1}(3))}\]

So, we’ll need $f’$, which we can easily get since we know $f$, and $f^{-1}(3)$. For the latter, we don’t know the function for $f^{-1}$ but we can find it by solving $f(x)=3$.

Quick note about solving $f(x)=3$ and any similar step in similar problems. The book glosses over how they get their solution. In the above example from the book, they just draw the rest of the owl and say it’s 2, and also do the same for any similar worked out exercises.

Typically, you’ll be given some other piece of information to help you solve these, but you can also do a mix of solving and guessing. In this case, $\frac{1}{4}x^3 + x - 1=3$ cleans up to $x(x^2 + 4) = 16$. Checking your factors of 16 will hopefully have you arrive at a solution of 2.

After some complicated math, we get $f(2)=3$, so $f^{-1}(3)=2$. Now we can run that through the derivative of $f$.

\[f'(2) = \frac{3}{4}(2)^2+1 = 4\]

So the finished product is

\[(f^{-1})'(3) = \frac{1}{f'(f^{-1}(3))} = \frac{1}{f'(2)} = \frac{1}{4}\]

Another property of inverse functions and their derivatives is that there slopes are also inverses. We found that $(f^{-1})’(3) = \frac{1}{4}$, but we also saw that $f’(2)=4$. The slope of of the tangent line at one point multiplied by it’s inverse’s slope, will always get you a 1.

Derivatives of Inverse Trigonometric Functions

As usual, there is another list of derivative rules. Arcsine and arctangent are the popular ones, but keep in mind their matching pair is just negated. We’ll go through the proof of arcsine in class.