Question

What is the derivative of `[tan(abs(x))]^(-1)`?

Answer 1

`d/dx(cot|x|)={(-csc^2x, x>=0),(csc^2x, x< 0):}`

Explanation:

We have `(tan|x|)^-1=cot|x|` `{(cotx, x>=0),(-cotx, x <0):}`

Note that `cot(-x)=cos(-x)/sin(-x)=cosx/-sinx=-cotx`

`cotx=cosx/sinx`

`d/dx(cotx)=d/dx(cosx/sinx)`

`d/dx(cosx/sinx)=(-sin^2x-cos^2x)/sin^2x=-1/sin^2x=-csc^2x`

`therefored/dx(-cotx)=csc^2x`

`d/dx(cot|x|)={(-csc^2x, x>=0),(csc^2x, x< 0):}`