How do you differentiate $y=sin^-1sqrtx$ ?

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Answer 1 · 2016-11-23T19:05:48

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

$y= sin^-1sqrtx <=> siny=x^(1/2)$

Differentiating (Implicitly) wrt $x$ :

$cosydy/dx = 1/2x^(-1/2)$
$:. cosydy/dx = 1/(2sqrtx)$

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

And so:
$:. sqrt(1 -x)dy/dx = 1/(2sqrtx)$
$:. dy/dx = 1/(2sqrtxsqrt(1 -x))$

How do you differentiate y=sin^-1sqrtxy=sin^-1sqrtx?