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

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Answer 1 · 2017-02-22T20:25:09

See below

$f(x)=(ln(x^2-1))/x^2$

For $F(x)=lnf(x)$ , $F'(x)=(f'(x))/f(x)$

For $F(x)=(g(x))/(h(x))$ , $F'(x)=(g(x)h'(x)-h(x)g'(x))/(h(x))^2$

Thus $f'(x)=((2x^2)/(x^2-1)-2xln(x^2-1))/x^4=$

$((2x)/(x^2-1)-2ln(x^2-1))/x^3$

Cleaning this up, we get:

$f'(x)=(2x-2ln(x^2-1)^(x^2-1))/(x^3(x^2-1))$

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