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

Jun 29, 2015

#### Answer:

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

#### Explanation:

I would start by using a rule of the logarithms to take the $\frac{1}{2}$ in front of the log:
$y = \frac{1}{2} \ln \left(1 - {x}^{2}\right)$

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:

$y ' = \frac{1}{2} \left[\frac{1}{1 - {x}^{2}} \cdot \left(- 2 x\right)\right] = \frac{x}{{x}^{2} - 1}$