Question

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

Answer 1

Derivative of `ln(1-x^2)` is `-(2x)/(1-x^2)`

Explanation:

We use te concept of function of a function. If we have `f(g(x))`, then

`(df)/(dx)=(df)/(dg(x))xx(dg)/(dx)`

Here we can write `f(x)=ln(1-x^2)` as `f(x)=ln(g(x))`, where `g(x)=1-x^2`

Now as `f(x)=ln(g(x))`, `(df)/(dg(x))=1/(g(x))`

and `(dg)/(dx)=-2x` and hence

`(df)/(dx)=1/(g(x))xx(-2x)=(-2x)/(g(x))=-(2x)/(1-x^2)`