Chapter 3.5, Problem 53E

(a)

To determine

To show: The derivative of the inverse function ${\left(f^{-1}\right)}^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(f^{-1}\left(x\right)\right)}$.

Explanation of Solution

Given:

The function $f\left(x\right)$ is one to one and differentiable and its inverse $f^{-1}$ also differentiable.

Derivative rules:

Chain rule: $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

Proof:

Obtain the derivative of the function $f^{-1}\left(x\right)$.

Since the function $f\left(x\right)$ is one to one function, $f\left(f^{-1}\left(x\right)\right)=x$.

Let $u=f^{-1}\left(x\right)$.

$f\left(u\right)=x$

Differentiate implicitly with respect to x.

$\begin{array}{c}\frac{d}{dx}\left(f\left(u\right)\right)=\frac{d}{dx}\left(x\right)\\ \frac{d}{dx}\left(f\left(u\right)\right)=1\end{array}$

Apply the chain rule and simplify the terms,

$\begin{array}{l}\frac{d}{du}\left(f\left(u\right)\right)\frac{du}{dx}=1\\ f^{\prime }\left(u\right)\frac{du}{dx}=1\end{array}$

Substitute $u=f^{-1}\left(x\right)$,

$f^{\prime }\left(f^{-1}\left(x\right)\right)\frac{d}{dx}\left(f^{-1}\left(x\right)\right)=1$

Divided by $f^{\prime }\left(f^{-1}\left(x\right)\right)$ on both sides,

$\begin{array}{l}\frac{d}{dx}\left(f^{-1}\left(x\right)\right)=\frac{1}{f^{\prime }\left(f^{-1}\left(x\right)\right)}\\ {\left(f^{-1}\right)}^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(f^{-1}\left(x\right)\right)}\end{array}$

Hence the required result is proved.

(b)

To determine

To find: The value $f^{-1}\left(5\right)$.

Answer to Problem 53E

The value of ${\left(f^{-1}\right)}^{\prime }\left(5\right)$ is $\frac{3}{2}$.

Explanation of Solution

Given:

The values $f\left(4\right)=5\text{ and }{f}^{\prime }\left(4\right)=\frac{2}{3}$

Calculation:

Obtain the value of ${\left(f^{-1}\right)}^{\prime }\left(5\right)$.

From part (a), ${\left(f^{-1}\right)}^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(f^{-1}\left(x\right)\right)}$.

Substitute 5 for x,

${\left(f^{-1}\right)}^{\prime }\left(5\right)=\frac{1}{f^{\prime }\left(f^{-1}\left(5\right)\right)}$

Since the function is one to one, $f\left(4\right)=5$ implies that $f^{-1}\left(5\right)=4$.

Substitute $f^{-1}\left(5\right)=4$ and $f^{\prime }\left(4\right)=\frac{2}{3}$,

$\begin{array}{c}{\left(f^{-1}\right)}^{\prime }\left(5\right)=\frac{1}{f^{\prime }\left(4\right)}\\ =\frac{1}{\frac{2}{3}}\\ =\frac{3}{2}\end{array}$

Therefore, the value of ${\left(f^{-1}\right)}^{\prime }\left(5\right)$ is $\frac{3}{2}$.