Question
Find the second-order derivative of the function x3 log x
$\therefore \frac{{dy}}{{dx}} = {x^3}\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}{x^3}$
$= {x^3}.\frac{1}{x} + \log x\left( {3{x^2}} \right)$
= x2 + 3x2log x
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {{x^2} + 3{x^2}\log x} \right)$
$= \frac{d}{{dx}}{x^2} + 3\frac{d}{{dx}}\left( {{x^2}\log x} \right)$
$= 2x + 3\left[ {{x^2}\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}{x^2}} \right]$
$= 2x + 3\left( {{x^2}.\frac{1}{x} + \left( {\log x} \right)2x} \right)$
= 2x + 3(x + 2x log x)
= 2x + 3x + 6x log x
= 5x + 6x log x
= x(5 + 6 log x)
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| Differential equation | Function |
| $\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$ | $\text{y}=\pm\sqrt{\text{a}^2-\text{x}^2}$ |