Question

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

Answer 1

`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!