Question

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

Answer 1

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

Explanation:

You could use the quotient rule, but I typically avoid doing this whenever possible as I find it tends to lead to higher chance of making an error and is generally more strenuous. To differentiate using the product rule, rewrite as

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

Product rule:

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

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

In our case, `g(x)=x` and `h(x)=(x-1)^-1`

Leaving `g(x)` alone and multiply by the derivative of `h(x)`, for which we would use the chain rule.

`h'(x)=-(x-1)^-2*1`

Where `1` is the derivative of the inside term, `x-1`.

Then, we leave `h(x)` alone and multiply by `g'(x)`

`g'(x)=1`

Putting it all together, we have

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

Which is equivalent to

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

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

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