How do you differentiate #f(x) = sin(2x)cos(2x)# using the product rule?

Question

Charles Arocha · Accepted Answer

See the explanation section below.

Differentiate #f(x) = sin(2x)cos(2x)# using the product rule I use the order: the derivative of a product of functions is the derivative of the first times the second, plus: the first times the derivative of the second. #d/dx(FS) = F'S+FS'# Note that we shall need the chain rule for the derivatives of #sin(2x)# and #cos(2x)# You may choose to write #F#, #S#, #F'# and #S'# before using the formula. I do not have that habit, so #f'(x) = [cos(2x)(2)]cos(2x)+sin(2x)[-sin(2x)(2)]# # = 2cos^2(2x)-2sin^2(2x)# Your teacher/textbook may well prefer to rewrite this answer using #cos(2theta) = cos^2theta - sin^2theta#, to get # = 2[cos^2(2x)-sin^2(2x)]# # = 2cos(4x)# Although if you're going to do that, I suggest Rewriting the function Use #sin(2theta) = 2sintheta costheta# to rewrite #f(x)# as #f(x) = sin(2x)cos(2x) = 1/2sin(4x)#. Now we do not need the product rule, only the chain rule (which we needed in the other method also).

Axel Alvarez · Answer

undefined To differentiate $ f(x) = \sin(2x)\cos(2x) $ using the product rule, you apply the formula $ (uv)' = u'v + uv' $. First, find the derivatives of $ \sin(2x) $ and $ \cos(2x) $.

$ \frac{d}{dx}[\sin(2x)] = 2\cos(2x) $
$ \frac{d}{dx}[\cos(2x)] = -2\sin(2x) $

Then, apply the product rule:

$ f'(x) = (\sin(2x))(-2\sin(2x)) + (2\cos(2x))(\cos(2x)) $

Simplify the expression:

$ f'(x) = -2\sin^2(2x) + 2\cos^2(2x) $