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

1 Answer
Nov 1, 2016

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#