# How do you find the derivative of y=xln^3x?

Aug 19, 2017

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

#### Explanation:

We need to start with the product rule. Where $u$ and $v$ are both functions, if $y = u v$, then:

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

Thus, where $y = x {\ln}^{3} x$:

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

Now we have two internal derivatives we need to figure out. The first is basic: $\frac{d}{\mathrm{dx}} x = 1$.

For the second derivative, we need the chain rule. First, note that we have a function cubed: ${\ln}^{3} x = {\left(\ln x\right)}^{3}$. Through the power rule, $\frac{d}{\mathrm{dx}} {x}^{3} = 3 {x}^{2}$, but if there were a more complex function instead of $x$ we see that $\frac{d}{\mathrm{dx}} {u}^{3} = 3 {u}^{2} \frac{\mathrm{du}}{\mathrm{dx}}$--that is, we still do the power rule but then multiply by the derivative of the inner function.

Thus, $\frac{d}{\mathrm{dx}} {\ln}^{3} x = 3 {\ln}^{2} x \left(\frac{d}{\mathrm{dx}} \ln x\right)$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 \cdot {\ln}^{3} x + x \left(3 {\ln}^{2} x\right) \left(\frac{d}{\mathrm{dx}} \ln x\right)$

Recall that $\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\ln}^{3} x + 3 x {\ln}^{2} x \left(\frac{1}{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\ln}^{3} x + 3 {\ln}^{2} x$

If you wish:

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