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

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