Question

How do you find the derivative of `cotx`?

Answer 1

`dy/dx = -csc^2x`

Explanation:

`y = cotx`

`y = 1/tanx`

`y = 1/(sinx/cosx)`

`y = cosx/sinx`

Letting `y= (g(x))/(h(x))`, we have that `g(x) = cosx` and `h(x) = sinx`.

`y' = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2`

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

`y' = (-sin^2x - cos^2x)/(sinx)^2`

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

`y' = -1/sin^2x`

`y' = -csc^2x`

Hopefully this helps!