How do you differentiate #y=tanx-cotx#?

1 Answer
Noah G
Dec 8, 2016

#y' = -(4(sin(2x) - cos(2x)))/sin^2(2x)#

Explanation:

Convert to sine and cosine:

#y = sinx/cosx - cosx/sinx#

Differentiate using the quotient rule--twice.

#y' = (cosx xx cosx - (-sinx xx sinx))/(cosx)^2 - (-sinx xx sinx - (cosx xx cosx))/(sinx)^2#

#y'= (cos^2x +sin^2x)/cos^2x - (-sin^2x - cos^2x)/sin^2x#

#y' = 1/cos^2x + (cos^2x + sin^2x)/sin^2x#

#y' = 1/cos^2x + 1/sin^2x#

#y' = sec^2x + csc^2x#

Hopefully this helps!

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