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.

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

Short answer

(a) $f^{\text{'}}\left(c\right)=1$

(b)$f^{\text{'}}\left(c\right)=1$

Step-by-step solution

Part (a) Step 1. Given information.

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

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,

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

Part (b) Step 1. Find f'(c)

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

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