Question

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

Answer 1

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

Explanation:

We can use the Quotient Rule to find the derivative, which states:

`y'=(g(x)h'(x)-h(x)g'(x))/(g(x))^2`

Where in our case,

`h(x)=x` and `g(x) = x-1`

We then need to find the derivative of each of these:

`h'(x) = 1` and `g'(x)=1`

Reassembling this we find our derivative:

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

Answer 2

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

Explanation:

In a previous answer, we used the quotient rule which is a simplification of the product rule. Some people, including me, don't like remembering more than one rule, so let's do that again using the product rule , which states:

`f'(x) = g'(x)h(x) + g(x)h'(x)`

in our case

`g(x) = x` and `h(x) = (x-1)^(-1)`

Now we need to find the derivatives of these functions:

`g'(x) = 1` and `h'(x) = -(x-1)^(-2)`

So reassembling we find our solution:

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