How do you differentiate y=tanx-cotx?
1 Answer
Dec 8, 2016
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!