Chapter 3.1, Problem 63E

(a)

To determine

To show : the derivative of $\cos \left(x°\right)$ is $-\frac{\pi }{180}\sin \left(x°\right)$ .

Explanation of Solution

Given information :

The function is $\cos \left(x°\right)$ .

Formula needed :

Chain rule of derivative: $\frac{d}{dx}\left(f\left(v\right)\right)=f^{\prime }\left(v\right)\cdot \frac{dv}{dx}$ .

  $1°=\frac{\pi }{180}$

Differentiate $\cos \left(x°\right)$ with respect to $x$ ,

  $\begin{array}{c}\frac{d}{dx}\cos \left(x°\right)=-\sin \left(x°\right)\cdot \frac{d}{dx}\left(x°\right)\\ =-\sin \left(x°\right)\cdot \frac{d}{dx}\left(\frac{x\pi }{180}\right)\\ =-\sin \left(x°\right)\cdot \frac{\pi }{180}\\ =-\frac{\pi }{180}\sin \left(x°\right)\end{array}$

Hence proved.

(b)

To determine

To find : the derivative of $\tan \left(x°\right)$ .

Answer to Problem 63E

  $\frac{d}{dx}\tan \left(x°\right)=\frac{\pi }{180}{\sec }^2\left(x°\right)$

Explanation of Solution

Given information :

The function is $\tan \left(x°\right)$ .

Formula needed :

Chain rule of derivative: $\frac{d}{dx}\left(f\left(v\right)\right)=f^{\prime }\left(v\right)\cdot \frac{dv}{dx}$ .

  $1°=\frac{\pi }{180}$

Differentiate $\tan \left(x°\right)$ with respect to $x$ ,

  $\begin{array}{c}\frac{d}{dx}\tan \left(x°\right)={\sec }^2\left(x°\right)\cdot \frac{d}{dx}\left(x°\right)\\ ={\sec }^2\left(x°\right)\cdot \frac{d}{dx}\left(\frac{x\pi }{180}\right)\\ ={\sec }^2\left(x°\right)\cdot \frac{\pi }{180}\\ =\frac{\pi }{180}{\sec }^2\left(x°\right)\end{array}$

The derivative of $\tan \left(x°\right)$ is $\frac{\pi }{180}{\sec }^2\left(x°\right)$ .