Question

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

Answer 1

Recall that `d/dxln(x)=1/x`.

So, by the chain rule, when we put a function inside the natural log function, we see that its derivative is

`d/dxln(f(x))=1/f(x)*f'(x)`

So here,

`dy/dx=d/dxln(1-x^2)=1/(1-x^2)[d/dx(1-x^2)]=(-2x)/(1-x^2)`

If you want, you can simplify the negative/minus signs:

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