How do you differentiate #f(x)= sin2xcos2x# using the product rule?

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Answer 1 · 2017-07-19T14:51:09

# f'(x) = 2cos4x #

Using the trig identity:

# sin2A -= 2sinAcosA #

We have:

# sin4A -= 2sin2Acos2A #

So we can write:

# f(x) = sin2xcos2 #
# " " = 1/2sin4x #

And so, using the chain rule we have:

# f'(x) = 1/2cos4x * 4#
# " " = 2cos4x #

This is an easier method than using the product rule