# How do you find the second derivative of  ln^2 (x)?

Jul 11, 2016

$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 \left(1 - \ln x\right)}{x} ^ 2 = \ln \left\{{\left(\frac{e}{x}\right)}^{\frac{2}{x} ^ 2}\right\} .$

#### Explanation:

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln x \cdot \frac{d}{\mathrm{dx}} \left(\ln x\right)$.......[Chain Rule].....$= \frac{2 \ln x}{x} .$

$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{d}{\mathrm{dx}} \left\{\frac{\mathrm{dy}}{\mathrm{dx}}\right\}$........................[Defn.]
$= \frac{d}{\mathrm{dx}} \left\{\frac{2 \ln x}{x}\right\}$
$= 2 \left[\frac{x \cdot \frac{d}{\mathrm{dx}} \left(\ln x\right) - \left(\ln x\right) \cdot \frac{d}{\mathrm{dx}} \left(x\right)}{x} ^ 2\right]$..........[Quotient Rule]
$= 2 \left[\frac{x \cdot \frac{1}{x} - \left(\ln x\right) \left(1\right)}{x} ^ 2\right]$
$= \frac{2 \left(1 - \ln x\right)}{x} ^ 2$
$= \left(\frac{2}{x} ^ 2\right) \left(\ln e - \ln x\right)$
$= \left(\frac{2}{x} ^ 2\right) \cdot \ln \left(\frac{e}{x}\right)$
$= \ln \left\{{\left(\frac{e}{x}\right)}^{\frac{2}{x} ^ 2}\right\}$

Maths. is amazing, enjoy it!