How do you differentiate #f(x)=(2x^3)(ln3x)(sinx) # using the product rule?
1 Answer
Aug 15, 2017
# f'(x) = 6x^2 \ ln3x \ sinx + 2x^2 \ sinx + 2x^3 \ ln3x \ cosx #
Explanation:
We have:
# f(x)=(2x^3)(ln3x)(sinx) #
We need to use the triple product rule, giving
# f'(x) = (d/dx2x^3)(ln3x)(sinx) + (2x^3)(d/dxln3x)(sinx) + (2x^3)(ln3x)(d/dxsinx) #
# " " = (6x^2)(ln3x)(sinx) + (2x^3)(1/(3x)d/dx3x)(sinx) + (2x^3)(ln3x)(cosx) #
# " " = (6x^2)(ln3x)(sinx) + (2x^3)((3)/(3x))(sinx) + (2x^3)(ln3x)(cosx) #
# " " = 6x^2 \ ln3x \ sinx + 2x^2 \ sinx + 2x^3 \ ln3x \ cosx #