How do you find the derivative of $arcsin(sqrt(1-x^2))$ ?

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Answer 1 · 2016-11-29T23:19:04

$d/dx arcsin(sqrt(1-x^2))= -x/(|x|sqrt(1-x^2))$

Let $y = arcsin(sqrt(1-x^2))$
Then $siny = sqrt(1-x^2) = (1-x^2)^(1/2)$

Differentiating Implicitly and applying the chain rule;
$cosydy/dx = (1/2)(1-x^2)^(-1/2)(-2x)$
$:. cosydy/dx = -x/sqrt(1-x^2) ... [1]$

Then using the fundamental trig identity $sin^2A+cos^2A-=1$ we have:

$cos^2y=1-sin^2y$
$cos^2y=1-((1-x^2)^(1/2))^2$
$:. cos^2y=1 - (1-x^2))$
$:. cos^2y=x^2$
$:. cosy=+-x$
$:. cosy=|x|$

Substituting into [1] we get:
$:. |x|dy/dx = -x/sqrt(1-x^2)$

Leading to the solution:
$dy/dx = -x/(|x|sqrt(1-x^2))$

How do you find the derivative of arcsin(sqrt(1-x^2)) arcsin(sqrt(1-x^2)) ?