How do you differentiate #f(x)=csc3x * cos2x# using the product rule?

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Answer 1 · 2016-01-02T05:12:55

#f'(x)=-csc3x(3cot3xcos2x+2sin2x)#

The product rule states that

#d/dx[f(x)g(x)]=f'(x)g(x)+g'(x)f(x)#

Thus,

#f'(x)=cos2xd/dx(csc3x)+csc3xd/dx(cos2x)#

Find each derivative, using the chain rule both times. Recall that

#d/d(cscu)=-u'cscucotu# and #d/dx(cosu)=-u'sinu#

Therefore,

#d/dx(csc3x)=-3csc3xcot3x#

and

#d/dx(cos2x)=-2sin2x#

Plug these back in.

#f'(x)=-3csc3xcot3xcos2x-2sin2xcsc3x#

#=-csc3x(3cot3xcos2x+2sin2x)#