# How do you differentiate x(lnx)^2?

Jun 9, 2016

Use the product rule and simplify.
$\frac{d}{\mathrm{dx}} \left[x \cdot {\ln}^{2} \left(x\right)\right] = \ln \left(x\right) \left[\ln \left(x\right) + 2\right]$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left[x \cdot {\ln}^{2} \left(x\right)\right]$

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

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

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

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

$= \ln \left(x\right) \left[\ln \left(x\right) + 2\right]$