Worksheet for Section 3.6

Secton 3.6 is about inverse functions, and in particular how to find the derivative of an inverse function (if you know the derivative of the original function). Recall that $g(x)$ is an inverse function for $f(x)$ if $f(g(x))=x$ and $g(f(x))=x$. If $f(x)$ has such an inverse on its domain, then the inverse is unique -- the normal notation is to write $g(x)=f^{-1}(x)$. Note that the domain of $f^{-1}(x)$ may not be the same as the domain of $f(x)$. Start with a couple of examples -- to begin, let $f(x)=x^2$ with domain restricted to $x\geq 0$, so that $g(x)=f^{-1}(x)=\sqrt{x}$. Graph both functions, and locate the point $(2,4)$ on the graph of $f(x)$, and the corresponding point $(4,2)$ on the graph of $f^{-1}(x)$. Now find the slopes of the tangent lines to the graphs at these points. To see this relationship another way, note that $[g(x)]^2=x$, and take derivatives on both sides (on the left side, use the General Power Rule, and leave $g'(x)$ in your expression). Now solve for $g'(x)$. For the next example, begin by noting that $e^{\ln x}=x$ and take the derivatives on both sides again, using the Chain Rule on the left side but leaving the derivative of $\ln x$ in your expression. Now solve for the derivative of $\ln x$. Here is the resulting rule for the derivative of $\ln x$:

$$\frac{d}{dx}\ln x=\frac1x$$

These examples suggest a reciprocal relationship between the derivatives of $f(x)$ and $f^{-1}(x)$. Here are the key facts relating continuity and differentiability of $f^{-1}(x)$ to the same properties of the original function $f(x)$:

• if $f(x)$ is continuous on its domain, and $f(x)$ has inverse function $f^{-1}(x)$, then $f^{-1}(x)$ is continuous on its domain
• if $f(x)$ is differentiable at $c$ and $f'(c)\neq 0$, then $f^{-1}(x)$ is differentiable at $f(c)$
• if $f(x)$ is differentiable in an interval $I$, and $f(x)$ has a differentiable inverse function $f^{-1}(x)$, then $f^{-1}(x)$ is differentiable at any $x$ for which $f'(f^{-1}(x))\neq 0$, and $\frac{d}{dx}f^{-1}(x)$ can be computed as follows:
  $$\frac{d}{dx}f^{-1}(x)=\frac1{f'(f^{-1}(x))}$$
  This is a direct consequence of the Chain Rule, proceding as in the examples above from the equation $f(f^{-1}(x))=x$ and taking derivatives on both sides.

As another example, for the function $h(x)=\frac12x^3-1$, compute $h^{-1}(x)$ and $(h^{-1})'(x)$ when $x=3$, both without actually finding a formula for $h^{-1}(x)$. You do have to worry to some extent about whether a function has an inverse -- for example, unless you restrict the domain as in the example above, the function $f(x)=x^2$ does not have an inverse. (Why not?)

This technique for computing the derivative of an inverse function provides formulas for the derivatives of inverse trigonometric functions:

$$\frac{d}{dx}[\arcsin x]= \frac{1}{\sqrt{1-x^2}}\qquad \frac{d}{dx... ...cos x]= \frac{-1}{\sqrt{1-x^2}}\qquad \frac{d}{dx}[\arctan x]= \frac{1}{1+x^2}$$

(I will show you in class how to derive the arcsine formula, using both the above formula for the derivative of an inverse function and implicit differentiation.) Here are a few problems involving these inverse trigonometric functions -- in each case, compute the indicated derivative:

$$\frac{d}{dx}[\arcsin(e^{3x})]\qquad\qquad \frac{d}{dx}[\arccos(3x)]$$

$$\frac{d}{dx}[\arctan(\sqrt{x})]\qquad\qquad \frac{d}{dx}[\arcsin(x)+x\sqrt{1-x^2}]$$