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

Oct 15, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4}{x \ln \left(2 {x}^{4}\right)}$

#### Explanation:

$y = \ln \left(\ln \left(2 {x}^{4}\right)\right)$

Through the chain rule, we see that:

$\frac{d}{\mathrm{dx}} \ln \left(\textcolor{red}{f} \left(x\right)\right) = \frac{1}{\textcolor{red}{f}} \left(x\right) \cdot \frac{d}{\mathrm{dx}} \textcolor{red}{f} \left(x\right)$

So, here, we see that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \ln \left(\textcolor{red}{\ln} \left(2 {x}^{4}\right)\right) = \frac{1}{\textcolor{red}{\ln}} \left(2 {x}^{4}\right) \cdot \frac{d}{\mathrm{dx}} \textcolor{red}{\ln} \left(2 {x}^{4}\right)$

Note that we will use the same rule again to find $\frac{d}{\mathrm{dx}} \ln \left(2 {x}^{4}\right)$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} \left(2 {x}^{4}\right) \cdot \left[\frac{d}{\mathrm{dx}} \ln \left(\textcolor{red}{2 {x}^{4}}\right)\right] = \frac{1}{\ln} \left(2 {x}^{4}\right) \cdot \left[\frac{1}{\textcolor{red}{2 {x}^{4}}} \cdot \frac{d}{\mathrm{dx}} \textcolor{red}{2 {x}^{4}}\right]$

Differentiating $2 {x}^{4}$ through the power rule gives $8 {x}^{3}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} \left(2 {x}^{4}\right) \cdot \frac{1}{2 {x}^{4}} \cdot 8 {x}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} \left(2 {x}^{4}\right) \cdot \frac{1}{x} \cdot 4$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4}{x \ln \left(2 {x}^{4}\right)}$