Question

Find the derivatives of$f\left(x\right)=\frac{1}{x^2}.$

Short answer

The derivative of $f\left(x\right)=\frac{1}{x^2}is-\frac{2}{x^3}.$

Step-by-step solution

Given Information 

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

Simplification 

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \\ f\text{'}\left(\frac{1}{x^2}\right)=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{{(x+h)}^2}}-{\displaystyle \frac{1}{x^2}}}{h} \\ =\underset{h\to 0}{\lim }\frac{x^2-x^2-2hx-h^2}{x^2{(x+h)}^{2}h} \\ =\underset{h\to 0}{\lim }\frac{-(2x+h)h}{{(x^2+hx)}^{2}h} \\ =\underset{h\to 0}{\lim }\frac{-2x-h}{{(x^2+hx)}^2} \\ =\frac{-2x-0}{{(x^2+h.0)}^2} \\ =\frac{-2x}{x^4} \\ =-\frac{2}{x^3} \end{gathered}$$

Therefore, the derivate of$f\left(x\right)=\frac{1}{x^2}is-\frac{2}{x^3}.$