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

1 Answer
Jun 29, 2015

Answer:

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)#