How do you differentiate #f(x)=cos5x * cot3x# using the product rule?

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Answer 1 · 2016-07-16T06:09:36

#-5sin5xcot3x-3csc^2(3x)cos5x#

The derivative of a product is stated as follows:
#color(blue)((u(x)*v(x))'=u'(x)*v(x)+v'(x)*u(x))#

Take #u(x)=cos(5x)# and #v(x)=cot(3x)#
Let's find #u'(x)# and #v'(x)#

Knowing the derivative of trigonometric function that says:
#(cosy)'=-y'siny# and
#(cot(y))' =-y'(csc^2y)#
So,
#u'(x)=(cos5x)'=-(5x)'sin5x=-5sin5x#

#v'(x)=(cot3x)'=-(3x)'csc^2(3x)=-3csc^2(3x)#

Thus,
#color(blue)(f'(x)=(u(x)*v(x))')#

Substituting #u'(x)# and #v'(x)# in the above property we have:

#=-5sin5xcot3x-3csc^2(3x)cos5x#