How do you find the derivative of #f(x)= 2x sin(x) cos(x)#?

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Answer 1 · 2016-08-02T20:21:40

# f'(x) = sin(2x) + 2xcos(2x)#

We're going to be using the product rule anyway, but instead of having to deal with 3 terms we can simplify this by noticing that

#2sin(x)cos(x) = sin(2x)#

So #f(x)=xsin(2x)#

#f'(x) = d/(dx)(x)sin(2x) + xd/(dx)(sin(2x))#

First term, derivative of #x# is #1#.

We differentiate #sin(2x)# using the chain rule.

#u = 2x implies (du)/(dx) = 2#

#y = sin(u) implies (dy)/(du) = cos(u)#

#implies (dy)/(dx) = 2cos(2x)#

#therefore f'(x) = sin(2x) + 2xcos(2x)#