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

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Answer 1 · 2015-06-29T21:46:43

I found: $y'=x/(x^2-1)$
(I assumed that only the argument of the log is raised to the power $1/2$ )

I would start by using a rule of the logarithms to take the $1/2$ in front of the log:
$y=1/2ln(1-x^2)$

Now I would use the Chain Rule to derive the log first as it is and multiply by the derivative of the argument to get:

$y'=1/2[1/(1-x^2)*(-2x)]=x/(x^2-1)$

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