# How do you differentiate y = (ln(x^2))^(2x+3)?

Jun 14, 2018

Note firstly that $y$ can be expressed more simply.
$y = {\left(\ln \left({x}^{2}\right)\right)}^{2 x + 3}$
$y = {\left(2 \ln x\right)}^{2 x + 3}$

Take the variable out of the exponent by taking logarithms:
$\ln y = \left(2 x + 3\right) \ln \left(2 \ln x\right)$

Differentiate with the product rule (RHS) and the chain rule (LHS):
$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln \left(2 \ln x\right) + \left(2 x + 3\right) \frac{d}{\mathrm{dx}} \left[\ln \left(2 \ln x\right)\right]$

Differentiate the double log by the chain rule:
$\frac{d}{\mathrm{dx}} \left[\ln \left(2 \ln x\right)\right] = \frac{1}{2 \ln x} \cdot \frac{d}{\mathrm{dx}} \left[2 \ln x\right] = \frac{1}{2 \ln x} \cdot \frac{2}{x} = \frac{1}{x \ln x}$

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

Thus
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(2 \ln x\right)}^{2 x + 3} \left[2 \ln \left(2 \ln x\right) + \frac{2 x + 3}{x \ln x}\right]$
or, expressed in the fashion of the question
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\ln \left({x}^{2}\right)\right)}^{2 x + 3} \left[2 \ln \ln \left({x}^{2}\right) + \frac{2 x + 3}{x \ln x}\right]$

Jun 14, 2018

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

#### Explanation:

We seek the derivative of:

$y = {\left(\ln \left({x}^{2}\right)\right)}^{2 x + 3}$

We can take Natural Logarithms of both sides:

$\ln y = \ln \left\{{\left(\ln \left({x}^{2}\right)\right)}^{2 x + 3}\right\}$

And using the logarithm properties, $\ln {a}^{b} = b \ln a$ and $\ln a b = \ln a + \ln b$, this becomes:

$\ln y = \left(2 x + 3\right) \ln \left\{\ln \left({x}^{2}\right)\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus = \left(2 x + 3\right) \ln \left\{2 \ln x\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus = \left(2 x + 3\right) \left\{\ln 2 + \ln \left(\ln x\right)\right\}$

Then by applying the product rule, and differentiating implicitly, we have:

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

Applying the chain rule we get:

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

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{2 x + 3}{x \ln x} + 2 \ln 2 + 2 \ln \left(\ln x\right)$

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