# How do you find the first and second derivative of ln(lnx^2)?

Aug 18, 2016

$\left(1\right) : \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x \ln x}$.

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

#### Explanation:

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

Using the well-known Rules of Log. Fun., we have,

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

By the Chain Rule, then,

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

$= 0 + \frac{1}{\ln} x \cdot \frac{d}{\mathrm{dx}} \left(\ln x\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} x \cdot \frac{1}{x} = \frac{1}{x \ln x}$.

Before proceeding further for the ${2}^{n d}$ Derivative $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$, let us recall that $\frac{d}{\mathrm{dt}} \left(\frac{1}{t}\right) = - \frac{1}{t} ^ 2$. Hence,

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

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

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

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

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