# What is the derivative of ln(4x)?

Jun 29, 2015

1/x

#### Explanation:

$y = \ln 4 x$

We have a choice. We can either use the chain rule in the form:
$\frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) = \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ OR we can use properties of logarithms to rewrite the function.

Chain Rule Solution

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

Rewrite Solution

Use $\ln a b = \ln a + \ln b$, to get:

$\frac{d}{\mathrm{dx}} \left(\ln 4 x\right) = \frac{d}{\mathrm{dx}} \left(\ln 4 + \ln x\right) = \frac{d}{\mathrm{dx}} \left(\ln 4\right) + \frac{d}{\mathrm{dx}} \left(\ln x\right) = 0 + \left(\frac{1}{x}\right) = \frac{1}{x}$

(Note that $\ln 4$ is some constant, hence its derivative is $0$.)