Logarithmic Differentiation
Sometimes it's just easier to take the derivative of
$\ln(y)$ than of $y$. In those cases, we can get $y'$ indirectly by using the chain rule. In fact,
$$ \frac{d}{dx} \Bigl(\ln(y)\Bigr)= \frac{\frac{dy}{dx}}{\quad y\quad },$$
or, equivalently
$$\frac{dy}{dx} = y \frac{d}{dx}\Bigl(\ln(y)\Bigr).$$
So, to compute $\displaystyle \frac{dy}{dx}$, first compute $\ln(y)$, then take its
derivative, and then multiply the answer by $y$. This procedure isn't
always helpful, and in some cases it can make computations much
harder, but in others it can make them much easier.
Example: Find the derivative of $y=x^x$.
This is a classic example: we can't take the derivative of
$x^x$ from the product, quotient or chain rules (at least not without
some serious tricks), so it looks like we're stuck. However, $$\ln(y)
= x\ln(x),$$ which we do know how to
differentiate. By the product rule, the derivative of $x
\ln(x)$ is $\displaystyle (x)\Bigl(\frac{1}{x}\Bigr)+(1)\left(\ln(x)\right)=1+\ln(x)$, so that
$$
\frac{d}{dx}\Bigl( x^x \Bigr)=
x^x \left(1+\ln(x)\right).$$

When should you use logarithmic differentiation?
— Whenever $\ln(y)$
is easier to differentiate than $y$! If $y$ involves a bunch of
powers, products and quotients, then $\ln(y)$ is likely to be simpler,
thanks to the three basic rules of logs:
$$\ln(ab) = \ln(a)+\ln(b),
\qquad \ln\left(\frac{a}{b}\right)=\ln(a)\ln(b), \qquad \ln(a^r) = r \ln(a).$$
On the other hand, if $y$ is a sum of terms, then $\ln(y)$ is likely to be a
mess, since there's no simple rule for $\ln(a+b)$. In other words,
your mileage will vary.
