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, $$ \frac{d}{dx} \ln(y)=
\frac{1}{y}\frac{dy}{dx},$$ or, equivalently $$\frac{dy}{dx} = y
\frac{d}{dx}\left(\ln(y)\right).$$
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 $y=x^x$.
DO: Try to work this
problem before reading further.
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), nor the rule for differentiating exponential
functions, 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 $y$. This is the case when you have
$y=(f(x))^{g(x)}$ for functions $f$ and $g$.
