# What are the first and second derivatives of f(x)=ln(x^(2x+1) ) ?

Nov 27, 2015

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

2nd derivative: $\frac{2}{x} + \frac{1}{{x}^{2}}$ Did not have time to do the $\frac{{d}^{2}}{\mathrm{dx}}$ properly so just gave the answer!

#### Explanation:

Given:$y = \ln \left({x}^{2 x + 1}\right)$

Write as : $y = \left(2 x - 1\right) \ln \left(x\right)$

Using standard form $\frac{\mathrm{dy}}{\mathrm{dx}} = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$
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Let $u = 2 x - 1 \to \frac{\mathrm{du}}{\mathrm{dx}} = 2$
Let $v = \ln \left(x\right) \to \textcolor{w h i t e}{\ldots} \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x}$

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

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

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