Question

How do you differentiate `f(x)=1/(x^3-1)*x^2*cosx` using the product rule?

Answer 1

`f'(x)=-(x^4+2x)/(x^3-1)^2*cos(x)-x^2/(x^3-1)*sin(x)`

Explanation:

Using the product rule

`(uv)'=u'v+uv'` we get with
`u=x^2/(x^3-1)`

`u'=(2x(x^3-1)-x^2*3x^2)/(x^3-1)^2`
simplifying we get

`u'=-(x^4+2x)/(x^3-1)^2`
`v=cos(x)`
`v'=-sin(x)`

putting thins together we get

`f'(x)=-(x^4+2x)/(x^3-1)^2cos(x)-x^2/(x^3-1)sin(x)`