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

Aug 22, 2016

y'=[2{x^2-(x^2-1)ln(x^2-1)}]/(x^3(x^2-1).

$y ' ' = \frac{2}{x} ^ 2 \left\{\frac{3 \ln \left({x}^{2} - 1\right)}{x} ^ 2 - \frac{{x}^{2} + 1}{{x}^{2} - 1} ^ 2\right\}$.

#### Explanation:

Let $y = \frac{\ln \left({x}^{2} - 1\right)}{x} ^ 2. \ldots \ldots . \left(1\right)$

$\therefore {x}^{2} y = \ln \left({x}^{2} - 1\right) = \ln \left\{\left(x + 1\right) \left(x - 1\right)\right\}$

${x}^{2} y = \ln \left(x + 1\right) + \ln \left(x - 1\right)$.

Diff. ing both sides w.r.t. $x$, we get,

${x}^{2} \cdot y ' + y \left({x}^{2}\right) ' = \left(\ln \left(x + 1\right)\right) ' + \left(\ln \left(x - 1\right)\right) '$.

$\therefore {x}^{2} y ' + y \left(2 x\right) = \left\{\frac{1}{x + 1}\right\} \left(x + 1\right) ' + \left\{\frac{1}{x - 1}\right\} \left(x - 1\right) '$.

$\therefore {x}^{2} y ' + 2 x y = \frac{1}{x + 1} + \frac{1}{x - 1} = \frac{2 x}{{x}^{2} - 1} \ldots \ldots \ldots . \left(2\right)$

$\therefore y ' = \frac{1}{x} ^ 2 \left[2 x \left\{\frac{1}{{x}^{2} - 1} - y\right\}\right]$, or, using $\left(1\right)$ for $y$,

$y ' = \frac{2}{x} \left\{\frac{1}{{x}^{2} - 1} - \frac{\ln \left({x}^{2} - 1\right)}{x} ^ 2\right\}$

Hence, y'=[2{x^2-(x^2-1)ln(x^2-1)}]/(x^3(x^2-1).

For $y ' '$, we rewrite $\left(2\right)$, in the following more useful form :

${x}^{2} y ' + 2 x y = \frac{1}{x + 1} + \frac{1}{x - 1}$.

Diff.ing, both sides w.r.t. $x$, we get,

${x}^{2} y ' ' + y ' \left(2 x\right) + 2 \left(x y ' + y\right) = - \frac{1}{x + 1} ^ 2 - \frac{1}{x - 1} ^ 2$

$\therefore {x}^{2} y ' ' + 4 x y ' + 2 y = - \left\{\frac{{\left(x - 1\right)}^{2} + {\left(x + 1\right)}^{2}}{{\left(x + 1\right)}^{2} {\left(x - 1\right)}^{2}}\right\}$

$\therefore {x}^{2} y ' ' + 4 x y ' + 2 y = - \frac{2 \left({x}^{2} + 1\right)}{{x}^{2} - 1} ^ 2$

Here, $x y ' = 2 \left\{\frac{1}{{x}^{2} - 1} - y\right\}$

$\therefore {x}^{2} y ' ' + 4 \cdot 2 \left\{\frac{1}{{x}^{2} - 1} - y\right\} + 2 y = - \frac{2 \left({x}^{2} + 1\right)}{{x}^{2} - 1} ^ 2$

$\therefore {x}^{2} y ' ' + \frac{8}{{x}^{2} - 1} - 8 y + 2 y = - \frac{2 \left({x}^{2} + 1\right)}{{x}^{2} - 1} ^ 2$

$\therefore {x}^{2} y ' ' - 6 y = - \frac{2 \left({x}^{2} + 1\right)}{{x}^{2} - 1} ^ 2$

$\therefore {x}^{2} y ' ' = 6 y - \frac{2 \left({x}^{2} + 1\right)}{{x}^{2} - 1} ^ 2 = 2 \left\{3 y - \frac{{x}^{2} + 1}{{x}^{2} - 1} ^ 2\right\}$

$= 2 \left\{\frac{3 \ln \left({x}^{2} - 1\right)}{x} ^ 2 - \frac{{x}^{2} + 1}{{x}^{2} - 1} ^ 2\right\}$

Finally, $y ' ' = \frac{2}{x} ^ 2 \left\{\frac{3 \ln \left({x}^{2} - 1\right)}{x} ^ 2 - \frac{{x}^{2} + 1}{{x}^{2} - 1} ^ 2\right\}$.

Enjoy Maths.!