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.

$f\left(x\right)=e^x,x=0$

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$f\left(x\right)=e^x$

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,

$\begin{array}{r}\underset{h\to 0}{lim} \frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim} \frac{e^{0+h}-e^0}{h}\\ =\underset{h\to 0}{lim} \frac{e^h-1}{h}\\ =\underset{h\to 0}{lim} \frac{\left(1+\frac{h}{1!}+\frac{h^2}{2!}+\cdots \right)-1}{h}\\ =\underset{h\to 0}{lim} \frac{\frac{h}{1!}+\frac{h^2}{2!}+\cdots }{h}\\ =\underset{h\to 0}{lim} \frac{h\left(\frac{1}{1!}+\frac{h}{2!}+\cdots \right)}{h}\\ =\underset{h\to 0}{lim} \frac{1}{1!}+\frac{h}{2!}+\cdots \\ =1\end{array}$

Part (b) Step 1. Find f'(c)

Find the derivate of the function using x→0 definition

Therefore,

$\begin{array}{r}\underset{x\to 0}{lim} \frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0}{lim} \frac{e^x-e^0}{x}\\ =\underset{x\to 0}{lim} \frac{e^x-1}{x}\\ =\underset{x\to 0}{lim} \frac{\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots \right)-1}{x}\\ =\underset{x\to 0}{lim} \frac{x\left(1+\frac{x}{2!}+\cdots \right)}{x}\\ =\underset{x\to 0}{lim} \left(1+\frac{x}{2!}+\cdots \right)\\ =1\end{array}$