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: $sin(y) = x$

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

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