Question

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

Answer 1

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

Explanation:

The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So if `f(x) = 1/(x-1)` then;

And so the derivative of `f(x)` is given by:

`f'(x) = lim_(h rarr 0) ( 1/((x+h)-1) - 1/(x-1) ) /h`
`\ \ \ \ \ \ \ \ = lim_(h rarr 0) ( (x-1)-(x+h-1) ) /( h(x-1)(x+h-1)`
`\ \ \ \ \ \ \ \ = lim_(h rarr 0) ( -h ) /( h(x-1)(x+h-1)`
`\ \ \ \ \ \ \ \ = lim_(h rarr 0) - (1 ) /((x-1)(x+h-1)`
`\ \ \ \ \ \ \ \ = - (1 ) /((x-1)(x+0-1)`
`\ \ \ \ \ \ \ \ = - (1 ) /((x-1)^2`