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?