How do you find the derivative of #f(x) = xcot(x)#?

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Answer 1 · 2018-04-04T08:27:09

#cot(x)-xcsc^2(x)#

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

And so, we let #f(x)=x,g(x)=cot(x)#.

Therefore, #f'(x)=1,g'(x)=-csc^2(x)#.

We get:

#d/dx(xcot(x))=1*cot(x)+x*(-csc^2(x))#

#=cot(x)-xcsc^2(x)#