Question

Use the definition of the derivative to find f'(x) if
$f(x) = \frac{1}{x - 1}$
​

Answer 1

Answer:

f'(x) = - $\frac{1}{(x-1)^2}$

Step-by-step explanation:

f'(x) = $lim_{h>0}$ $\frac{f(x+h)-f(x)}{h}$

      = $lim_{h>0}$ $\frac{\frac{1}{x+h-1}-\frac{1}{x-1} }{h}$

     = $lim_{h>0}$ $\frac{x-1-(x+h-1)}{h(x+h-1)(x-1)}$

     = $lim_{h>0}$ $\frac{x-1-x-h+1}{h(x+h-1)(x-1)}$

     = $lim_{h>0}$ $\frac{-h}{h(x+h-1)(x-1)}$ ← cancel h on numerator denominator

     = - $\frac{1}{(x-1)^2}$