How do you find the second derivative of $y= ln(1-x^2)^(1/2)$ ?

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Answer 1 · 2016-10-30T00:54:52

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

$y = ln(1-x^2)^(1/2)$ can be rewritten as $y^2=ln(1-x^2)$

Differentiating wrt $x$ gives:
$d/dx(y^2) = d/dx( ln(1-x^2) )$

$:. dy/dxd/dy(y^2) = d/dx{ ln(1-x^2) }$

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

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

$:. ln(1-x^2)^(1/2)(dy/dx) = (x)/(x^2-1)$

$:. dy/dx = (x)/((x^2-1)sqrtln(1-x^2))$

How do you find the second derivative of y= ln(1-x^2)^(1/2) y= ln(1-x^2)^(1/2) ?