Question

Find the derivative of the following functions from first principle
$\frac{1}{x^2}$

Answer 1

Let $f\left(x\right)=\frac{1}{x^2}$
Thus according to first principle,
$f^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{\frac{1}{(x+h)}-\frac{1}{x^2}}{h}$
$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{x-{(x+h)}^2}{x^2{(x+h)}^2}\right]$
$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{x^2-x^2-h^2-2hx}{x^2{(x+h)}^2}\right]$
$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{-h^2-2hx}{x^2{(x+h)}^2}\right]$
$=\underset{h\to 0}{lim}\left[\frac{-h-2x}{x^2{(x+h)}^2}\right]$
$=\frac{0-2x}{x^2{(x+0)}^2}=\frac{-2}{x^3}$