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

2 Answers
Topscooter
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)#.

Lucio Falabella
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))#

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