Question

Find the derivative of the following function from first principle:
$\frac{x+1}{x-1}$.

Answer 1

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