How do you differentiate f(x)=x^2 * sin4x using the product rule?

2 Answers
Jan 7, 2016

f'(x) = 2xsin(4x) + 4x^2cos(4x)

Explanation:

By the product rule, the derivative of u(x)v(x) is u'(x)v(x) + u(x)v'(x). Here, u(x) = x^2 and v(x) = sin(4x) so u'(x) = 2x and v'(x) = 4cos(4x) by the chain rule.

We apply it on f, so f'(x) = 2xsin(4x) + 4x^2cos(4x).

Jan 7, 2016

f'(x)=2x*(sin(4x)+2xcos(4x))

Explanation:

Given a f(x)=h(x)*g(x) the rule is:

f'(x)=h'(x)*g(x)+h(x)*g'(x)

in this case:

h(x)=x^2
g(x)=sin(4x)

look at g(x) it is a composite function where the argoument is 4*x

g(x)=s(p(x))

then

g'(x)=s'(p(x))*p'(x)

d/dxf(x)=d/dxx^2*sin(4x)+x^2*d/dx sin(4x)*d/dx4x=

d/dxx^2*sin(4x)+x^2*d/dx sin(4x)*4d/dxx=

=2*x*sin(4x)+x^2*cos(4x)*4*1=

2x*sin(4x)+4x^2cos(4x)=2x*(sin(4x)+2xcos(4x))