# Activity7.2Derivatives of Inverse Functions¶ permalink

Using Definition 3.3.1, it is easy to establish that $\lzoo{x}{x^2}=2x\text{.}$ We can use this formula and implicit differentiation to find the formula for $\lzoo{x}{\sqrt{x}}\text{.}$ If $y=\sqrt{x}\text{,}$ 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 $y=\sqrt{x}\text{,}$ 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 $\lzoo{x}{\fe{\sin^{-1}}{x}}\text{.}$ If $y=\fe{\sin^{-1}}{x}\text{,}$ then $\fe{\sin}{y}=x$ and $-\frac{\pi}{2}\leq y\leq \frac{\pi}{2}\text{.}$ 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 $\fe{\cos}{y}=\sqrt{1-\fe{\sin^2}{y}}\text{.}$ 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 $\lzoo{x}{\fe{\ln}{x}}=\frac{1}{x}\text{.}$ 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 $x\text{.}$

##### 2

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

##### 3

Use the fact that $\lzoo{x}{\fe{\tan}{x}}=\fe{\sec^2}{x}$ together with implicit differentiation to show that $\lzoo{x}{\fe{\tan^{-1}}{x}}=\frac{1}{1+x^2}\text{.}$ 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 $x\text{.}$ Please note that you will need to use the Pythagorean identity that relates the tangent and secant functions while working this problem.