How do you find the derivative of #y=sin^-1x#?

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Answer 1 · 2016-02-10T10:58:10

#dy/dx = 1/sqrt(1-x^2)#

This can usually be quoted off of a table of standard derivatives:

#d/dxsin^(-1)x=1/sqrt(1-x^2)#

However, if you have trouble convincing yourself then a good place to start would be implicit differentiation:

We know that if # y = sin^-1(x)# then
#sin(y) = x#

Applying implicit differentiation we get:

#cos(y)dy/dx = 1#

#->dy/dx =1/ cos(y)#

Now use the trig identity: #cos^2(y)+sin^2(y) = 1# to get:

#dy/dx = 1/sqrt(1-sin^2(y))#

And from the start we know that: #sin(y) = x#

#therefore dy/dx = 1/sqrt(1-x^2)#