Question

What is the second derivative of `x/(x-1)` and the first derivative of `2/x`?

Answer 1

Question 1
If `f(x) = (g(x))/(h(x))` then by the Quotient Rule

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

So if `f(x) = x/(x-1)`
then the first derivative
`f'(x) = ((1)(x-1) - (x)(1))/x^2`

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

and the second derivative is
`f''(x) = 2x^-3`

Question 2
If `f(x) = 2/x` this can be re-written as

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

and using standard procedures for taking the derivative
`f'(x) = -2x^-2`
or, if you prefer
`f'(x) = - 2/x^2`