Chapter 3.3, Problem 6E

To determine

To find : derivative of $y$ with respect to $s$ .

Answer to Problem 6E

The derivative of $y=s\sqrt{1-s^2}+{\cos }^{-1}s$ is $-\frac{s^2}{2\sqrt{1-s^2}}+\sqrt{1-s^2}-\frac{1}{s}$ .

Explanation of Solution

Given information :

A function $y=s\sqrt{1-s^2}+{\cos }^{-1}s$ .

Formula needed :

  $\cos x=\sqrt{1-{\sin }^{2}x}$ and $\frac{d}{dx}\left(\sin u\right)=\cos u\cdot \frac{du}{dx}$

Since, $y=s\sqrt{1-s^2}+{\cos }^{-1}s$ .

Assume, $t={\cos }^{-1}s$

So, $y=s\sqrt{1-s^2}+t$

Multiply by $\cos$ on both sides of $t={\cos }^{-1}s$ ,

  $\cos t=s$

Differentiate above with respect to $s$ ,

  $\begin{array}{c}\frac{d}{ds}\cos t=\frac{d}{ds}\left(s\right)\\ -\sin t\cdot \frac{dt}{ds}=1\\ \frac{dt}{ds}=-\frac{1}{s}\end{array}$

Differentiate $y=s\sqrt{1-s^2}+t$ with respect to $s$ ,

  $\begin{array}{c}\frac{dy}{ds}=\frac{d}{ds}\left(s\sqrt{1-s^2}+t\right)\\ \frac{dy}{ds}=\frac{d}{ds}\left(s\sqrt{1-s^2}\right)+\frac{d}{ds}\left(t\right)\\ \frac{dy}{ds}=s\frac{d}{ds}\left(\sqrt{1-s^2}\right)+\sqrt{1-s^2}\frac{d}{ds}\left(s\right)+\frac{d}{ds}\left(t\right)\\ \frac{dy}{ds}=s\frac{1}{2\sqrt{1-s^2}}\frac{d}{ds}\left(1-s^2\right)+\sqrt{1-s^2}+\frac{dt}{ds}\end{array}$

Differentiate and simplify,

  $\begin{array}{c}\frac{dy}{ds}=s\frac{1}{2\sqrt{1-s^2}}\frac{d}{ds}\left(1-s^2\right)+\sqrt{1-s^2}+\frac{dt}{ds}\\ \frac{dy}{ds}=s\frac{1}{2\sqrt{1-s^2}}\left(-2s\right)+\sqrt{1-s^2}+\frac{dt}{ds}\\ \frac{dy}{ds}=-\frac{s^2}{2\sqrt{1-s^2}}+\sqrt{1-s^2}+\frac{dt}{ds}\end{array}$

Substitute value of $\frac{dt}{ds}$ on $\frac{dy}{ds}=-\frac{s^2}{2\sqrt{1-s^2}}+\sqrt{1-s^2}+\frac{dt}{ds}$ .

  $\frac{dy}{ds}=-\frac{s^2}{2\sqrt{1-s^2}}+\sqrt{1-s^2}-\frac{1}{s}$

Thus, the derivative of $y=s\sqrt{1-s^2}+{\cos }^{-1}s$ is $-\frac{s^2}{2\sqrt{1-s^2}}+\sqrt{1-s^2}-\frac{1}{s}$ .