# Why is the derivative of ln -x = 1/x?

Mar 30, 2015

This is because the chain rule says:

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

So:

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

The $\frac{1}{-} x$ is the derivative of the logarithmic function and the $- 1$ is the derivative of $- x$.

For the same reason in this integral:

$\int \frac{1}{x} \mathrm{dx} = \ln | x | + c$

we have to put the absolue value to $x$, because we want to write all the primitive functions.