# Question #b5828

Oct 26, 2016

Ok, so we've got:

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

Now, let's transform this into:

$y = u \cdot v$

Whereby, u is a function of x and v is a function of x .

If this is the case:

$\frac{\mathrm{dy}}{\mathrm{dx}} = u \cdot \frac{\mathrm{dv}}{\mathrm{dx}} + v \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

This formula is actually called the product rule . Also, remember that y' is the same as $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Alright, so it turns out that:

$u = {x}^{2}$ which means that, $\frac{\mathrm{du}}{\mathrm{dx}} = 2 x$.

Also:

$v = \ln \left(2 x\right)$ which means that $\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x}$ thanks to the rule below:

When $y = \ln \left(f \left(x\right)\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right)}{f \left(x\right)}$.

With the information we've just produced, we can now figure out what $\frac{\mathrm{dy}}{\mathrm{dx}}$ will be...

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

$= x + 2 x \cdot \ln \left(2 x\right)$

*This result is y'.

Now, using a similar strategy, let's figure out what $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$ is. Remember that y'' is the same as $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$.

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

Which means that:

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

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

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

*This result is y''.

If you'd like to know how the product rule can be produced from scratch, visit the link below. It may help you understand why we use it.

https://socratic.org/calculus/basic-differentiation-rules/proof-of-the-product-rule/how-to-derive-the-product-rule-from-scratch