How do you differentiate #g(x) =x^2cotx# using the product rule?

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Answer 1 · 2016-01-02T16:26:56

#g'(x)=2xcotx-x^2csc^2x#

According to the product rule:

#g'(x)=cotxd/dx(x^2)+x^2d/dx(cotx)#

Find each derivative separately.

#d/dx(x^2)=2x#

and

#d/dx(cotx)=-csc^2x#

Plug these back in.

#g'(x)=2xcotx-x^2csc^2x#