Question

How do you find the derivative of `f(x)=1/(x-1)`?

Answer 1

`f'(x)=-1/(x-1)^2`

Explanation:

`f(x)=1/(x-1)=(x-1)^-1`

Using the rule: `f(x) =x^n => f'(x)=nx^(n-1)` and chain rule:

`f'(x)=-1*(x-1)^-2 * (x-1)' = -(x-1)^-2 = -1/(x-1)^2`

`f'(x)=-1/(x-1)^2`