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Activity7.2Derivatives of Inverse Functions

Using Definition 3.3.1, it is easy to establish that \(\lzoo{x}{x^2}=2x\). We can use this formula and implicit differentiation to find the formula for \(\lzoo{x}{\sqrt{x}}\). If \(y=\sqrt{x}\), then \(y^2=x\) (and \(y\geq0\)). Using implicit differentiation we have:\begin{align*} y^2&=x\\ \lzoo{x}{y^2}&=\lzoo{x}{x}\\ 2y\lz{y}{x}&=1\\ \lz{y}{x}&=\frac{1}{2y}\text{.} \end{align*}

But \(y=\sqrt{x}\), so we have:\(\begin{aligned}[t]\lz{y}{x}&=\frac{1}{2y}\\&=\frac{1}{2\sqrt{x}}\text{.}\end{aligned}\)

In a similar manner, we can use the fact that \(\lzoo{x}{\fe{\sin}{x}}=\fe{\cos}{x}\) to come up with a formula for \(\lzoo{x}{\fe{\sin^{-1}}{x}}\). If \(y=\fe{\sin^{-1}}{x}\), then \(\fe{\sin}{y}=x\) and \(-\frac{\pi}{2}\leq y\leq \frac{\pi}{2}\). Note that because of the angle range that \(y\) is confined to, that \(\fe{\cos}{y}\) is positive. So the Pythagorean identity \(\fe{\cos^2}{y}=1-\fe{\sin^2}{y}\) implies that \(\fe{\cos}{y}=\sqrt{1-\fe{\sin^2}{y}}\). This gives us:\begin{align*} \fe{\sin}{y}=x\\ \lzoo{x}{\fe{\sin}{y}}&=\lzoo{x}{x}\\ \fe{\cos}{y}\lz{y}{x}&=1\\ \lz{y}{x}&=\frac{1}{\fe{\cos}{y}}\\ \lz{y}{x}&=\frac{1}{\sqrt{1-\fe{\sin^2}{y}}}\\ \lz{y}{x}&=\frac{1}{\sqrt{1-x^2}}\text{.} \end{align*}

Subsection7.2.1Exercises

1

Use the fact that \(\lzoo{x}{e^x}=e^x\) together with implicit differentiation to show that \(\lzoo{x}{\fe{\ln}{x}}=\frac{1}{x}\). Begin by using the fact that \(y=\fe{\ln}{x}\) implies that \(e^y=x\) (and \(x>0\)). Your first step is to differentiate both sides of the equation \(e^y=x\) with respect to \(x\).

2

Use the fact that \(\lzoo{x}{\fe{\ln}{x}}=\frac{1}{x}\) together with implicit differentiation to show that \(\lzoo{x}{e^x}=e^x\). Begin by using the fact that \(y=e^x\) implies that \(\fe{\ln}{y}=x\). Your first step is to differentiate both sides of the equation \(\fe{\ln}{y}=x\) with respect to \(x\).

3

Use the fact that \(\lzoo{x}{\fe{\tan}{x}}=\fe{\sec^2}{x}\) together with implicit differentiation to show that \(\lzoo{x}{\fe{\tan^{-1}}{x}}=\frac{1}{1+x^2}\). Begin by using the fact that \(y=\fe{\tan^{-1}}{x}\) implies that \(\fe{\tan}{y}=x\) (and \(-\frac{\pi}{2}\lt y\lt\frac{\pi}{2}\)). Your first step is to differentiate both sides of the equation \(\fe{\tan}{y}=x\) with respect to \(x\). Please note that you will need to use the Pythagorean identity that relates the tangent and secant functions while working this problem.