# Question #7454e

Jul 19, 2017

${d}^{2} / {\mathrm{dx}}^{2} \left({x}^{3} \log x\right) = x \left(6 \log x + 5\right)$

#### Explanation:

Using the product rule for the second order derivative:

${d}^{2} / {\mathrm{dx}}^{2} \left({x}^{3} \log x\right) = \left({d}^{2} / {\mathrm{dx}}^{2} {x}^{3}\right) \log x + 2 \left(\frac{d}{\mathrm{dx}} {x}^{3}\right) \left(\frac{d}{\mathrm{dx}} \log x\right) + {x}^{3} \left({d}^{2} / {\mathrm{dx}}^{2} \log x\right)$

${d}^{2} / {\mathrm{dx}}^{2} \left({x}^{3} \log x\right) = 6 x \log x + 2 \times 3 {x}^{2} \times \frac{1}{x} + {x}^{3} \left(- \frac{1}{x} ^ 2\right)$

${d}^{2} / {\mathrm{dx}}^{2} \left({x}^{3} \log x\right) = 6 x \log x + 6 x - x$

${d}^{2} / {\mathrm{dx}}^{2} \left({x}^{3} \log x\right) = 6 x \log x + 5 x$

${d}^{2} / {\mathrm{dx}}^{2} \left({x}^{3} \log x\right) = x \left(6 \log x + 5\right)$