Question

What is the derivative of `y= ln(1-x^2)^(1/2)`?

Answer 1

`y=ln(1-x^2)^(1/2)`

`y^2=ln(1-x^2)` *To be continued...

Differentiate the left hand side of the equation using implicit differentiation. Differentiate the right hand side of the equation using the rule:

If `g(x)=ln(f(x))`, `g'(x)=(f'(x))/f(x)`

*Continued...

`2y*(dy)/(dx)=-(2x)/(1-x^2)`

`(dy)/(dx)=-(2x)/(1-x^2)*1/(2y)`

`(dy)/(dx)=-x/(1-x^2)*1/y`

`(dy)/(dx)=-x/((1+x)(1-x))*1/ln(1-x^2)^(1/2)`

`(dy)/(dx)=-x/(ln(1-x^2)^(1/2)*(1+x)(1-x))`

`(dy)/(dx)=-x/(ln((1+x)(1-x))^(1/2)*(1+x)(1-x))`

`(dy)/(dx)=-x/(sqrt(ln(1+x)+ln(1-x))*(1+x)(1-x))`

To tidy things up a bit, you can multiply the fraction to the right by one, which is the same as `(-1)/(-1)`.

So what you end up with is...

`(dy)/(dx)=x/(sqrt(ln(1+x)+ln(1-x))*(x-1)(x+1))`