Next: Working with other Bases Up: Doing that Calc Thing Previous: Integrating ex and its

Differentiating the Natural Log

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


\begin{displaymath}\frac{d}{dx} (\ln x) = \frac{1}{x} \end{displaymath}

Pretty amazing that the derivative of an ugly function like $\ln
x$ would be a pretty function like $\displaystyle \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 .

\begin{displaymath}\frac{d}{dx} \ln (\sqrt x) = \frac{1}{\sqrt x} \frac{1}{2\sqrt x} =
\frac{1}{2x} .\end{displaymath}


Option 2: Use the fact that

\begin{displaymath}\ln (\sqrt x) = \ln (x^ {\frac {1}{2}}) = \frac {1}{2} \ln x.\end{displaymath}

Then,

\begin{displaymath}\frac{d}{dx} \ln {( \sqrt x)} = \frac{d}{dx} (\frac {1}{2} \ln x ) =
\frac{1}{2x}.\end{displaymath}


In general, the chain rule  says

\begin{displaymath}\boxed{\frac{d}{dx} (\ln g(x)) = \frac{g'(x)}{g(x)}} .\end{displaymath}

So for instance,

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



Joel Hass
1999-05-26