Question

For the given function, find the derivative from the first principle. $\frac{x+1}{x-1}$

Answer 1

Solve for the derivation by first principle

Given function is, $\frac{x+1}{x-1}$

Let, $f\left(x\right)=\frac{x+1}{x-1}$

From the fundamental theorem of calculus, we know that,
$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Thus,
$\begin{array}{rcl}f\text{'}\left(x\right)& =& \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{\left(x+h\right)+1}{\left(x+h\right)-1}}-{\displaystyle \frac{x+1}{x-1}}}{h}\\ & =& \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{\left(x+h+1\right)\left(x-1\right)-\left[\left(x+1\right)\left(x+h-1\right)\right]}{\left(x+h-1\right)\left(x-1\right)}}}{h}\\ & =& \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{x^2+hx+\cancel{x}-h-\cancel{x}-1-\left[x^2+hx-\cancel{x}+\cancel{x}+h-1\right]}{\left(x+h-1\right)\left(x-1\right)}}}{h}\\ & =& \underset{h\to 0}{\lim }\frac{\cancel{x^2}+\cancel{hx}-h-\cancel{1}-\cancel{x^2}-\cancel{hx}-h+\cancel{1}}{h\left(x+h-1\right)\left(x-1\right)}\\ & =& \underset{h\to 0}{\lim }\frac{-2\cancel{h}}{\cancel{h}\left(x+h-1\right)\left(x-1\right)}\\ & =& \underset{h\to 0}{\lim }\frac{-2}{\left(x+h-1\right)\left(x-1\right)}\\ & =& -\frac{2}{\left(x+0-1\right)\left(x-1\right)}\\ & =& -\frac{2}{{\left(x+1\right)}^2}\end{array}$

Hence, the derivative of $\frac{x+1}{x-1}$ is $-\frac{2}{{\left(x+1\right)}^2}$.