How do you differentiate f'(x) = 2x log(x) + x?

Oct 14, 2015

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{2}{\ln \left(10\right)} + 2 \log \left(x\right) + 1$

Explanation:

We have, assuming that $\log \left(x\right)$ is the base 10 logarithm,

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

To find $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$ we need to use the product rule

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 2 x \frac{d}{\mathrm{dx}} \left(\log \left(x\right)\right) + \log \left(x\right) \frac{d}{\mathrm{dx}} \left(2 x\right) + \frac{d}{\mathrm{dx}} \left(x\right)$

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

We can rewrite $\log \left(x\right)$ as $\ln \frac{x}{\ln} \left(10\right)$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{2 x}{\ln} \left(10\right) \frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) + 2 \log \left(x\right) + 1$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{2 x}{x \ln \left(10\right)} + 2 \log \left(x\right) + 1$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{2}{\ln \left(10\right)} + 2 \log \left(x\right) + 1$

If you meant the natural log, just replace $\ln \left(10\right)$ for $\ln \left(e\right)$ or $1$