Q. 25 - Calculus

Question

Use (a) the $h\to 0$ definition of the derivative and then

(b) the $z\to c$ definition of the derivative to find $f\text{'}\left(c\right)$ for each function f and value $x=c$ in Exercises 23–38.

25. $f\left(x\right)=\frac{1}{x}, x=-1$

Step-by-Step Solution

Answer

$f\text{'}(-1)=-1$.

Part (a) Step 1. Given information.

A function is given as $f\left(x\right)=\frac{1}{x}$ and $x=c=-1$.

Part (a) Step 2. Find f ' ( - 1 ) using h → 0 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h} \\ f\text{'}(-1)=\underset{h\to 0}{\lim }\frac{f(-1+h)-f(-1)}{h} \\ =\underset{h\to 0}{\lim }\frac{\left({\displaystyle \frac{1}{h-1}}\right)-\left({\displaystyle \frac{1}{-1}}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{h-1}}+1}{h} \\ =\underset{h\to 0}{\lim }\frac{1+h-1}{h(h-1)} \\ =\underset{h\to 0}{\lim }\frac{h}{h(h-1)} \\ =\underset{h\to 0}{\lim }\frac{1}{h-1} \\ =\frac{1}{0-1} \\ =\frac{1}{-1} \\ =-1 \end{gathered}$$

Part (b) Step 1. Find f ' ( - 1 ) using z → - 1 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{z\to c}{\lim }\frac{f\left(z\right)-f\left(c\right)}{z-c} \\ f\text{'}(-1)=\underset{z\to -1}{\lim }\frac{f\left(z\right)-f(-1)}{z-(-1)} \\ =\underset{z\to -1}{\lim }\frac{{\displaystyle \frac{1}{z}}-\left({\displaystyle \frac{1}{-1}}\right)}{z+1} \\ =\underset{z\to -1}{\lim }\frac{{\displaystyle \frac{1}{z}}+1}{z+1} \\ =\underset{z\to -1}{\lim }\frac{1+z}{z(z+1)} \\ =\underset{z\to -1}{\lim }\frac{1}{z} \\ =\frac{1}{-1} \\ =-1 \end{gathered}$$