Question

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

Answer 1

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

Explanation:

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)`