Question

How to find formula for the nth derivative of `f(x)=x/((x-a)(x-b))`?

Answer 1

`d^n/(dx^n)(x/((x-a)(x-b))) = (-1)^n (n!)((a/(a-b))1/(x-a)^(n+1)-(b/(a-b))1/(x-b)^(n+1))`

Explanation:

Expanding

`x/((x-a)(x-b))=(a/(a-b))1/(x-a)-(b/(a-b))1/(x-b)`

now

`d^n/(dx^n)(1/(x-a))=((-1)^n n!)/(x-a)^(n+1)` so

`d^n/(dx^n)(x/((x-a)(x-b))) = (-1)^n (n!)((a/(a-b))1/(x-a)^(n+1)-(b/(a-b))1/(x-b)^(n+1))`