How do you find the derivative of #arcsin x + arccos x#?

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Answer 1 · 2017-01-02T00:53:06

#d/dx(arcsinx+arccosx)=0#

It may not at all be obvious but its worth remembering that:

#arcsinx+arccosx=pi/2#

Using this we have:

# d/dx(arcsinx+arccosx)=0#

If you could not spot this little helper then you can show the result using the individual results:

#d/(dx) arcsinx = 1/sqrt(1 - x^2) #
#d/(dx) arccosx = -1/sqrt(1 - x^2) #

And so the result is obvious