Question

Use the definition of the derivative to find $f\text{'}$for each function $f$in Exercises 34-59

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

Short answer

The value of$f\text{'}\left(x\right)=\frac{-2}{x^3}$

Step-by-step solution

Step 1. Given  information

The given function$f\left(x\right)=\frac{1}{x^2}$

Step 2. Finding the value of f'(x)

We know that $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}.........\left(1\right)$

Given $f\left(x\right)=\frac{1}{x^2}$the

Putting the values in (1)

localid="1648748944158" role="math" $$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{{(x+h)}^2}}-{\displaystyle \frac{1}{x^2}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{x^2-{(x+h)}^2}{x^2{(x+h)}^2}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{x^2-x^2-h^2-2xh}{x^2{(x+h)}^2}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{-h^2-2xh}{x^2{(x+h)}^2}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{h(-h-2x)}{x^2{(x+h)}^2}}}{h} \\ =\underset{h\to 0}{\lim }\frac{h(-h-2x)}{x^2{(x+h)}^2}\times \frac{1}{h} \\ =\underset{h\to 0}{\lim }\frac{(-h-2x)}{x^2{(x+h)}^2} \end{gathered}$$

Putting $h=0$

$$\begin{gathered} =\frac{-\left(0\right)-2x}{x^2.{(x+0)}^2} \\ =\frac{-2x}{x^2.x^2} \\ =\frac{-2}{x^3} \end{gathered}$$

Hence,the value of$f\text{'}\left(x\right)=\frac{-2}{x^3}$