Question

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

Short answer

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

Step-by-step solution

Given Information 

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

Simplification  

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

Therefore, the derivate of$\frac{1}{x}is-\frac{1}{x^2}.$