How do you differentiate y=tanx-cotx?

1 Answer
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!