Question

How do you differentiate `g(x) =x^2cosx` using the product rule?

Answer 1

`(dg(x))/(dx)=-x^2sinx+2xcosx`

Explanation:

`g(x)=x^2cosx` is product of two functions, say `f_1(x)=x^2` and `f_2(x)=cosx`.

Product rule states that if `g(x)=f_1(x)f_2(x)` then `g'(x)=f_1'(x)f_2(x)+f_1(x)f_2'(x)`.

Here `p'(x)` denotes differential of `p(x)` i.e.

`f_1'(x)` is differential of `f_1(x)` and `f_2'(x)` is differential of `f_2(x)`.

Hence, `(dg(x))/(dx)=x^2*(d(cosx))/dx+(d(x^2))/(dx)*cosx`

= `x^2*(-sinx)+2x*cosx=-x^2sinx+2xcosx`