How do you find the derivative of the function #f(x)=arcsinx+arccosx#?

1 Answer
Truong-Son N.
Jun 23, 2015

Notice the pattern between the two derivatives:

#d/(dx)[arcsinu] = 1/sqrt(1 - u^2)((du)/(dx))#

#d/(dx)[arccosu] = -1/sqrt(1 - u^2)((du)/(dx))#

Now you really only have to know one of them. #co# implies negative, then. It's also true for #d/(dx)[arc cotu]# vs. #d/(dx)[arc tanu]#, and #d/(dx)[arcsecu]# vs. #d/(dx)[arc cscu]#.

Now just do it.

#= 1/sqrt(1 - x^2) - 1/sqrt(1 - x^2) = 0#

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
1671 views around the world
You can reuse this answer
Creative Commons License