Q. 26 - 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.

26. $f\left(x\right)=\frac{1}{x^2}, x=2$

Step-by-Step Solution

Answer

$f\text{'}\left(2\right)=-\frac{1}{4}$.

Part (a) Step 1. Given information.

A function is given as $f\left(x\right)=\frac{1}{x^2}$ and $x=c=2$.

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

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h} \\ f\text{'}\left(2\right)=\underset{h\to 0}{\lim }\frac{f(2+h)-f\left(2\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{{(2+h)}^2}}-{\displaystyle \frac{1}{2^2}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{4+h^2+4h}}-{\displaystyle \frac{1}{4}}}{h} \\ =\underset{h\to 0}{\lim }\frac{4-4-4h-h^2}{4h(h^2+4h+4)} \\ =\underset{h\to 0}{\lim }\frac{-h(h+4)}{4h(h^2+4h+4)} \\ =\underset{h\to 0}{\lim }\frac{-(h+4)}{4(h^2+4h+4)} \\ =\frac{-(0+4)}{4(0+0+4)} \\ =\frac{-4}{16} \\ =-\frac{1}{4} \end{gathered}$$

Part (b) Step 1. Find f ' ( 2 ) using z → 2 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{z\to c}{\lim }\frac{f\left(z\right)-f\left(c\right)}{z-c} \\ f\text{'}\left(2\right)=\underset{z\to 2}{\lim }\frac{f\left(z\right)-f\left(2\right)}{z-2} \\ =\underset{z\to 2}{\lim }\frac{{\displaystyle \frac{1}{z^2}}-{\displaystyle \frac{1}{2^2}}}{z-2} \\ =\underset{z\to 2}{\lim }\frac{{\displaystyle \frac{1}{z^2}}-{\displaystyle \frac{1}{4}}}{z-2} \\ =\underset{z\to 2}{\lim }\frac{4-z^2}{4z^2(z-2)} \\ =\underset{z\to 2}{\lim }\frac{(2-z)(2+z)}{-4z^2(2-z)} \\ =\underset{z\to 2}{\lim }\frac{2+z}{-4z^2} \\ =-\frac{2+2}{4{\left(2\right)}^2} \\ =-\frac{4}{16} \\ =-\frac{1}{4} \end{gathered}$$