# What is the derivative of ln(2x)?

Apr 3, 2016

$\left(\ln \left(2 x\right)\right) ' = \frac{1}{2 x} \cdot 2 = \frac{1}{x} .$

#### Explanation:

You use the chain rule :

$\left(f \circ g\right) ' \left(x\right) = \left(f \left(g \left(x\right)\right)\right) ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$.

In your case : $\left(f \circ g\right) \left(x\right) = \ln \left(2 x\right) , f \left(x\right) = \ln \left(x\right) \mathmr{and} g \left(x\right) = 2 x$.

Since $f ' \left(x\right) = \frac{1}{x} \mathmr{and} g ' \left(x\right) = 2$, we have :

$\left(f \circ g\right) ' \left(x\right) = \left(\ln \left(2 x\right)\right) ' = \frac{1}{2 x} \cdot 2 = \frac{1}{x}$.

Apr 3, 2016

$\frac{1}{x}$

#### Explanation:

You can also think of it as

$\ln \left(2 x\right) = \ln \left(x\right) + \ln \left(2\right)$

$\ln \left(2\right)$ is just a constant so has a derivative of $0$.

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

Which gives you the final answer.