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

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Answer 1 · 2016-10-29T22:45:54

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

Let $y = (arcsinx)^2 => y^(1/2) = arcsinx$
$:. sin(y^(1/2)) = x$

Differentiating wrt $x$ ;
$:. cos(y^(1/2))(1/2y^(-1/2))dy/dx = 1$

$:. dy/dx = 2/(cos(y^(1/2))(y^(-1/2)))$

$:. dy/dx = (2y^(1/2))/(cos(y^(1/2)))$

Now $sin^2A + cos^2A -= 1 => cosA = sqrt(1 - sin^2A$
So, $cos(y^(1/2)) = sqrt(1 - sin^2(y^(1/2))$
$:. cos(y^(1/2)) = sqrt(1 - x^2)$

Hence, $dy/dx = (2arcsinx)/sqrt(1 - x^2)$

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