Now that we know the derivative of a log, we can combine it with
the chain rule:$$\frac{d}{dx}\Big( \ln(y)\Big)= \frac{1}{y}
\frac{dy}{dx}.$$ Sometimes it is easier to take the derivative of
$\ln(y)$ than of $y$, and it is the only way to differentiate some
functions. This is called logarithmic
differentiation.

The process of differentiating $y=f(x)$ with logarithmic
differentiation is simple. Take the natural log of both
sides, then differentiate both sides with respect to $x$.
Solve for $\frac{dy}{dx}$ and write $y$ in terms of $x$ and you
are finished.

Consider the function $(f(x))^{(g(x))}$, for any (non-constant)
functions $f$ and $g$. This is a
function for which we do not have a differentiation rule;
it is not a power function (because there is an $x$ in the
exponent), nor is it an exponential function (because there is an
$x$ in the base). Whenever you wish to
differentiate $(f(x))^{(g(x))}$, logarithmic differentiation works
beautifully. This is because once we take logs, we
can pull the power down and use the product rule.

Example: Find the derivative of $y=x^x$.

Solution: (Notice that both $f(x)=x$ and $g(x)=x$
above.) Take the log of both sides to get $\ln(y) = \ln(x^x)=x
\ln(x)$ (see how we can pull that exponent
$x$ down?). By the product rule,
the derivative of $x \ln(x)$ is $\ln(x) + 1$ (do
this work and simplify to get that answer). So we have
$$\frac{1}{y}\frac{dy}{dx}=\ln(x) + 1$$ and when we replace $y$ with
$x^x$ and solve for $\frac{dy}{dx}$ we get
$$
\frac{dy}{dx} = y \frac{d}{dx}\Big(\ln(y)\Big) = x^x (1+\ln(x)).
$$