How do you find the derivative of # y = cot5x + csc5x #?

1 Answer
Hariharan Baskarraj
May 13, 2018

#-5csc 5x (cosec 5x-cot 5x)or -5 csc 5x × tan 5x/2#

Explanation:

Given #y=cot 5x+csc 5x#
#d/dx(cot x)=-csc^2 x#
#d/dx(csc x)=-csc x cot x#

Now, By chain rule,
#d/dx(cot 5x)=-5csc^2 5x#
#d/dx(csc 5x)=-5csc 5x cot 5x#

Adding, we get,
#-5csc 5x (cosec 5x-cot 5x)# which is the answer.

Further simplifying,
#csc 5x-cot 5x " as " tan (5x/2)# by the identity #csc x-cot x =tan (x/2)#
We get the answer as#-5 csc 5x × tan 5x/2#

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