Question

How do you find the derivative of `arcsinx+arccosx`?

Answer 1

The derivative `=0`

Explanation:

First derivative of `arcsinx`
Let `u=arcsinx` `=>` `x=sinu`
take the derivative `1=((du)/dx)cosu`
Use the identity `cos^2u+sin^2u=1` `=>` `cosu=sqrt(1-x^2)`
So `(du)/dx=1/(sqrt(1-x^2))`
Let `v=arccosx` `=>` `x=cosv`
take the derivative `1=((dv)/dx)-sinv`
Use the identity `cos^2v+sin^2v=1` `=>` `sinv=sqrt(1-x^2)`
So `(dv)/dx=-1/(sqrt(1-x^2))`
Therefore `(arcsinx+arccosx)'=1/(sqrt(1-x^2))-1/(sqrt(1-x^2))=0`