# How do you find the derivative of ln(4x)?

Jan 21, 2017

It is $\frac{1}{x}$.

#### Explanation:

$\ln \left(4 x\right)$ is a composite function, composed of the functions $\ln x$ and $4 x$. Because of that, we should use the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

We already know that $\left(\ln x\right) ' = \frac{1}{x}$. So, we want what's inside of the natural logarithm to be a single variable, and we can do this by setting $u = 4 x$. Now we could say that $\left(\ln u\right) ' = \frac{1}{u}$, with respect to $u$. Essentially, the chain rule states that the derivative of $y$ with respect to $x$, is equal to the derivative of $y$ with respect to $u$, where $u$ is a function of $x$, times the derivative of $u$ with respect to $x$. In our case, $y = \ln \left(4 x\right)$. Differentiating $u$ with respect to $x$ is simple, since $u = 4 x$: $u ' = 4$, with respect to $x$. So, we see that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \cdot 4 = \frac{4}{u}$

Now we can change $u$ back into $4 x$, and get $\frac{4}{4 x} = \frac{1}{x}$.

Interestingly enough, $\left[\ln \left(c x\right)\right] '$ where $c$ is a non-zero constant, where it is defined, is equal to $\frac{1}{x}$, just like $\left(\ln x\right) '$, even though we are using the chain rule.