Question

How do you differentiate `f(x)=(x+3sinx)(x^2-3x)` using the product rule?

Answer 1

`f'(x) = (x+3sinx)(2x-3) + (x^2-3x)(1+3cosx)`

Explanation:

You need to use the product rule;
`d/dx(uv)=u(dv)/dx+v(du)/dx`

So with `f(x) = (x+3sinx)(x^2-3x)` we have;
`f'(x) = (x+3sinx)(d/dx(x^2-3x)) + (x^2-3x)(d/dx(x+3sinx))`

`:. f'(x) = (x+3sinx)(2x-3) + (x^2-3x)(1+3cosx)`

We could multiply out, and it may possibly simplify but the above is probably sufficient as an answer.