What is the derivative of this function $y=sin^-1(1-x)^(1/2)$ ?

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Answer 1 · 2016-11-30T00:03:58

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

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

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

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

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

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

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

What is the derivative of this function y=sin^-1(1-x)^(1/2)y=sin^-1(1-x)^(1/2)?