How do you differentiate #f(x)=cot(5x)+csc(5x)#?

1 Answer
Alan N.
Mar 2, 2017

#f'(x) = -5csc(5x)(csc(5x)+cot(5x))#

Explanation:

#f(x) = cot(5x) + csc(5x)#

#f'(x) = d/dx(cot(5x)) +d/dx(csc(5x))#

Applying standard differentials and the chain rule to each term:

#f'(x )# = #-csc^2 (5x) * d/dx(5x)# #- cot(5x) * csc(5x)# * #d/dx(5x)#

#= -csc^2 (5x) * 5 - cot(5x) * csc(5x) * 5#

#=-5csc(5x)(csc(5x)+cot(5x))#

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