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

Jun 28, 2017

The First Derivative is:

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

The Second Derivative is:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{1 + \ln x}{{x}^{2} {\ln}^{2} x}$

#### Explanation:

We have:

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

Using the law of logarithms we can write this as:

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

Then Differentiating wrt $x$ and applying the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 + \frac{1}{\ln} x \cdot \frac{1}{x}$
$\text{ } = \frac{1}{x \ln x}$

If we write this as:

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

Then we get the second derivative by differentiating again wrt $x$ and applying the chain and product rule:

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