How do you differentiate y=ln(9x+4)^2?

Oct 24, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{18 \ln \left(9 x + 4\right)}{9 x + 4}$

Explanation:

The chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$ for some function $u$ of $x$.

In this problem, let $u = \ln \left(9 x + 4\right)$, so $y = {u}^{2}$. Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \left[{u}^{2}\right] \times \frac{\mathrm{du}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 u \times \frac{\mathrm{du}}{\mathrm{dx}}$

And substitute back in for $u$.
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln \left(9 x + 4\right) \times \frac{\mathrm{du}}{\mathrm{dx}}$

So now we need to find $\frac{\mathrm{du}}{\mathrm{dx}}$, we can use the chain rule again. Now say $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dx}}$

Let $v = 9 x + 4$ so $u = \ln v$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dv}} \left[\ln v\right] \times \frac{\mathrm{dv}}{\mathrm{dx}}$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{v} \times \frac{\mathrm{dv}}{\mathrm{dx}}$

And substitute back in for $v$.
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{9 x + 4} \times \frac{\mathrm{dv}}{\mathrm{dx}}$

And finally we need to find $\frac{\mathrm{dv}}{\mathrm{dx}}$.
$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[9 x + 4\right]$
$\frac{\mathrm{dv}}{\mathrm{dx}} = 9$

And finally plug everything back in.
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{9 x + 4} \times 9 = \frac{9}{9 x + 4}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln \left(9 x + 4\right) \times \frac{9}{9 x + 4} = \frac{18 \ln \left(9 x + 4\right)}{9 x + 4}$