How do you differentiate #cos(3x)sin(3x)#?

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Answer 1 · 2016-11-12T15:20:39

See below.

You could use the product rule and the chain rule, but I find it more satisfying to rewrite the function before differentiating.

From trigonometry:

#sin(2theta) = 2sin(theta)cos(theta)#

Therefore

#sin(theta)cos(theta) = 1/2sin(2theta)#

So,

#f(x) = cos(3x)sin(3x) = 1/2sin(6x)#

Therefore,

#f'(x) = 1/2cos(6x) * d/dx(6x)# #" "# (chain rule)

# = 3cos(6x)#