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

1 Answer
Oct 29, 2015

Answer:

See the explanation section below.

Explanation:

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).