# How do you find the derivative of y=ln ln(3x^3)?

Nov 24, 2016

There are a couple of approaches, all of which will involve the chain rule, which I assume you are familiar with.

Method 1 - Use the chain rule (directly)

$y = \ln \left(\ln \left(3 {x}^{3}\right)\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln \left(3 {x}^{3}\right)} \cdot \frac{1}{3 {x}^{3}} \cdot 9 {x}^{2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{x \ln \left(3 {x}^{3}\right)}$

Method 2 - Use the chain rule (with substitution)

$\left\{\begin{matrix}\text{Let "u =3x^3 & => (du)/dx=9x^2 \\ "Let "v= lnu & => (dv)/(du)=1/u \\ "Then "y=lnv =ln(lnu)=ln(ln(3x^3))} & \implies \frac{\mathrm{dy}}{\mathrm{dv}} = \frac{1}{v}\end{matrix}\right.$

By the Chain Rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \cdot \frac{\mathrm{dv}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{v} \cdot \frac{1}{u} \cdot 9 {x}^{2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} u \cdot \frac{1}{3 {x}^{3}} \cdot 9 {x}^{2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} \left(3 {x}^{3}\right) \cdot \frac{3}{x}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{x \ln \left(3 {x}^{3}\right)}$

Method 3 - Take Exponents and differentiate Implicitly

$y = \ln \left(\ln \left(3 {x}^{3}\right)\right)$
$\therefore {e}^{y} = \ln \left(3 {x}^{3}\right)$
$\therefore {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3 {x}^{3}} \cdot 9 {x}^{2}$
$\therefore \ln \left(3 {x}^{3}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{x}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{x \ln \left(3 {x}^{3}\right)}$