What is the derivative of #sin^-1(x)#?

1 Answer
Jim H
Apr 15, 2016

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

Explanation:

Let #y=sin^-1x#,
so #siny=x# and #-pi/2 <= y <= pi/2# (by the definition of inverse sine).

Now differentiate implicitly:

#cosy dy/dx = 1#, so

#dy/dx = 1/cosy#.

Because #-pi/2 <= y <= pi/2#, we know that #cosy# is positive.

So we get:

#dy/dx = 1/sqrt(1-sin^2y) = 1/sqrt(1-x^2)#. (Recall from above #siny=x#.)

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