How do you differentiate #cos^2(3x)#?

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Answer 1 · 2016-12-10T22:22:35

If we think of #cos^2(x)# as our outside function, and #3x# as our inside function:

We would calculate as follows:

#d/dx cos^2(x)*d/dx(3x)#

Thus we would get:

#-2sin(3x)cos(3x)*3#

#=-6sin(3x)cos(3x)#