Q. 34 - 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. 

$f\left(x\right)=2e^x$

Step-by-Step Solution

Answer

(a) $f^{\text{'}}\left(c\right)=2$

(b) $f^{\text{'}}\left(c\right)=2$

Part (a) Step 1. Given information.

Given function is $f\left(x\right)=2e^x$

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,

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

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

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

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