# What is the derivative of ln(1/x)?

Apr 13, 2018

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

#### Explanation:

We could use the Chain Rule right away, but the properties of logarithms allow us to avoid that and make this easier.

$\ln \left(\frac{1}{x}\right) = \ln \left(1\right) - \ln x = - \ln x$

So,

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

Apr 13, 2018

$- \frac{1}{x}$

#### Explanation:

Well, let's try the chain rule, which states that,

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

Let $u = \frac{1}{x} , \therefore \frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{x} ^ 2$.

Then $y = \ln u , \therefore \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$.

Combining, we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \cdot - \frac{1}{x} ^ 2$

$= - \frac{1}{u {x}^{2}}$

Substituting back $u = \frac{1}{x}$, we get,

$= - \frac{1}{\frac{1}{x} \cdot {x}^{2}}$

$= - \frac{1}{{x}^{2} / x}$

$= - \frac{1}{x}$