# What are the first and second derivatives of  g(x) =x^2/(lnx)^2?

Nov 21, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \cdot \frac{\ln \left(x\right) - 1}{\ln} ^ 3 \left(x\right)$
$\frac{{\mathrm{dy}}^{2}}{\mathrm{dx}} ^ 2 = \frac{2 \left({\ln}^{2} \left(x\right) - 3 \ln \left(x\right) + 3\right)}{{\ln}^{4} \left(x\right)}$

#### Explanation:

We have

$y = {x}^{2} / {\left(\ln \left(x\right)\right)}^{2}$

We can use the product rule, the quotient rule, or the chain rule with logs. To make our life easier I'll use log derivativing.

Take the log of both sides

$\ln \left(y\right) = \ln \left({x}^{2} / {\left(\ln \left(x\right)\right)}^{2}\right)$

Use log properties

$\ln \left(y\right) = 2 \ln \left(x\right) - 2 \ln \left(\ln \left(x\right)\right)$

Derivate implicity

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{x} - \frac{2}{\ln} \left(x\right) \cdot \frac{1}{x}$

Multiply by $y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{\ln} ^ 2 \left(x\right) - \frac{2 x}{\ln} ^ 3 \left(x\right)$

Or

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \cdot \frac{\ln \left(x\right) - 1}{\ln} ^ 3 \left(x\right)$

To derivate again, do the same thing, now skipping over some steps

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

$\frac{\mathrm{dx}}{\mathrm{dy}} \cdot \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{2}{2 x} + \frac{1}{\ln \left(x\right) - 1} \cdot \frac{1}{x} - \frac{3}{\ln} \left(x\right) \cdot \frac{1}{x}$

Multiply both sides by $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{{\mathrm{dy}}^{2}}{\mathrm{dx}} ^ 2 = 2 \cdot \frac{\ln \left(x\right) - 1}{{\ln}^{3} \left(x\right)} + \frac{2 x}{\ln} ^ 3 \left(x\right) - 6 \cdot \frac{\ln \left(x\right) - 1}{\ln} ^ 4 \left(x\right)$

Or, using algebra to tie it all together

$\frac{{\mathrm{dy}}^{2}}{\mathrm{dx}} ^ 2 = \frac{2 \left({\ln}^{2} \left(x\right) - 3 \ln \left(x\right) + 3\right)}{{\ln}^{4} \left(x\right)}$