Differentiating the Natural Log

Let's not beat around the bush. Here you go:

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

Pretty amazing that the derivative of an ugly function like $\ln x$ would be a pretty function like $$\frac{1}{x}$$. It's kind of a Cinderella story for functions, but without the pumpkins, glass slipper and raging royal hormones.

Now, suppose that we want to differentiate $\ln {(\sqrt x)}$. We have two options:

Option 1: Just do it by the chain rule .

$$\frac{d}{dx} \ln (\sqrt x) = \frac{1}{\sqrt x} \frac{1}{2\sqrt x} = \frac{1}{2x} .$$

Option 2: Use the fact that

$$\ln (\sqrt x) = \ln (x^ {\frac {1}{2}}) = \frac {1}{2} \ln x.$$

Then,

$$\frac{d}{dx} \ln {( \sqrt x)} = \frac{d}{dx} (\frac {1}{2} \ln x ) = \frac{1}{2x}.$$

In general, the chain rule  says

$$\boxed{\frac{d}{dx} (\ln g(x)) = \frac{g'(x)}{g(x)}} .$$

So for instance,

$$\frac{d}{dx} (\ln(x^3-7)) =\frac{3x^2}{x^3-7}.$$