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

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Answer 1 · 2018-03-17T15:15:19

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

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

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

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