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

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