Q. 38 - Calculus

Question

Use (a) the h→0 definition of the derivative and then (b) the z→c definition of the derivative to find f'(c) for each function f and value x = c. 

$f\left(x\right)=\sec x,x=0$

Step-by-Step Solution

Answer

(a) $f^{\text{'}}\left(c\right)=0$

(b) $f^{\text{'}}\left(c\right)=0$

Part (a) Step 1. Given information.

Given function is $f\left(x\right)=\sec x$

We have to find $f^{\text{'}}\left(c\right)$

Part (a) Step 2. Find the f ' ( c )

We have to find the derivative of the function using h→0 definition,

Therefore,

$\begin{array}{r}\underset{h\to 0}{lim} \frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim} \frac{\sec (0+h)-\sec 0}{h}\\ =\underset{h\to 0}{lim} \frac{\sec h-1}{h}\\ =\underset{h\to 0}{lim} \frac{\frac{1}{\cos \left(h\right)}-1}{h}\\ =\underset{h\to 0}{lim} \frac{1-\cos \left(h\right)}{h\cos \left(h\right)}\\ =-\underset{h\to 0}{lim} \frac{\cos \left(h\right)-1}{h}\cdot \underset{h\to 0}{lim} \frac{1}{\cos \left(h\right)}\\ =-0\cdot \frac{1}{1}\\ =0\end{array}$

Part (b) Step 1. Find f ' ( c )

Find the derivate of the function using x→0 definition
Therefore,

$\begin{array}{r}\underset{x\to 0}{lim} \frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0}{lim} \frac{\sec x-\sec 0}{x}\\ =\underset{x\to 0}{lim} \frac{\sec x-1}{x}\\ =\underset{x\to 0}{lim} \frac{\frac{1-\cos x}{x}}{\cos }\\ =-\underset{x\to 0}{lim} \frac{1-\cos x}{x}\cdot \underset{x\to 0}{lim} \frac{1}{\cos x}\\ =-0\left(1\right)\\ =0\end{array}$