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

Question

1175 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-30T13:43:56

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

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):}#