Calculus of a Single Variable, Chapter 2, Review, Section Review, Problem 40

You need to evaluate the derivative of the given function, using the product tule for the products 3x*sin x and x^2*cos x , such that:
f'(x) = (3x)'*(sin x) + 3x*(sin x)' + (x^2)'(cos x) + (x^2)(cos x)'
f'(x) = 3*(sin x)+ 3x*(cos x) + 2x*cos x + x^2*(-sin x)
Combining like terms yields:
f'(x) = sin x*(3 - x^2) + x*(cos x)*(3 + 2)
f'(x) = sin x*(3 - x^2) + 5x*cos x
Hence, evaluating the derivative of the function, using the product rule where it is requested, yields f'(x) = sin x*(3 - x^2) + 5x*cos x.