Answer:
f'(x) = - [tex]\frac{1}{(x-1)^2}[/tex]
Step-by-step explanation:
f'(x) = [tex]lim_{h>0}[/tex] [tex]\frac{f(x+h)-f(x)}{h}[/tex]
= [tex]lim_{h>0}[/tex] [tex]\frac{\frac{1}{x+h-1}-\frac{1}{x-1} }{h}[/tex]
= [tex]lim_{h>0}[/tex] [tex]\frac{x-1-(x+h-1)}{h(x+h-1)(x-1)}[/tex]
= [tex]lim_{h>0}[/tex] [tex]\frac{x-1-x-h+1}{h(x+h-1)(x-1)}[/tex]
= [tex]lim_{h>0}[/tex] [tex]\frac{-h}{h(x+h-1)(x-1)}[/tex] ← cancel h on numerator denominator
= - [tex]\frac{1}{(x-1)^2}[/tex]
