You have formulas that allow you to differentiate $x^2\text{,}$ $2^x\text{,}$ and $2^2\text{.}$ You don't, however, have a formula to differentiate $x^x\text{.}$ In this section you are going to use a process called logarithmic differentiation to determine the derivative formula for the function $y=x^x\text{.}$ Example 7.3.1 shows this process for a different function.

With logarithmic differentiation, when $y$ is written as a function of $x\text{,}$ especially one involving only products, quotients, and powers, begin by taking the logarithm of both sides of the equation relating $y$ to $x\text{.}$ This technique will usually only be helpful if the logarithm rules allow you to break up the right side some. Proceed to differentiate each side with respect to $x\text{.}$ On the left side, respect the rules for implicit differentiation. Once this is done, algebraically solve for $\lz{y}{x}\text{.}$ If there are any instances of $y$ left, replace them with the original expression in $x\text{.}$

Example7.3.1Logarithmic Differentiation

\begin{align*} y&=\frac{x^{e^x}}{4x+1}\\ \fe{\ln}{y}&=\fe{\ln}{\frac{x^{e^x}}{4x+1}}\\ \fe{\ln}{y}&=\fe{\ln}{x^{e^x}}-\fe{\ln}{4x+1}\\ \fe{\ln}{y}&=e^x\fe{\ln}{x}-\fe{\ln}{4x+1}\\ \lzoo{x}{\fe{\ln}{y}}&=\lzoo{x}{e^x\fe{\ln}{x}-\fe{\ln}{4x+1}}\\ \frac{1}{y}\lz{y}{x}&=\lzoo{x}{e^x}\cdot\fe{\ln}{x}+e^x\cdot\lzoo{x}{\fe{\ln}{x}}-\frac{1}{4x+1}\cdot\lzoo{x}{4x+1}\\ \frac{1}{y}\lz{y}{x}&=e^x\cdot\fe{\ln}{x}+e^x\cdot\frac{1}{x}-\frac{1}{4x+1}\cdot4\\ \frac{1}{y}\lz{y}{x}&=e^x\fe{\ln}{x}+\frac{e^x}{x}-\frac{4}{4x+1}\\ \lz{y}{x}&=y\left(e^x\fe{\ln}{x}+\frac{e^x}{x}-\frac{4}{4x+1}\right)\\ \lz{y}{x}&=\frac{x^{e^x}}{4x+1}\left(e^x\fe{\ln}{x}+\frac{e^x}{x}-\frac{4}{4x+1}\right) \end{align*}

In the olden days (before symbolic calculators) we would use the process of logarithmic differentiation to find derivative formulas for complicated functions. The reason this process is “simpler” than straight forward differentiation is that we can obviate the need for the product and quotient rules if we completely expand the logarithmic expression before taking the derivative.

Subsection7.3.1Exercises

1

The function $y=x^x$ is only defined for positive values of $x$ (which in turn means $y$ is also positive), so we can say that $\fe{\ln}{y}=\fe{\ln}{x^x}\text{.}$ Use implicit differentiation to find a formula for $\lz{y}{x}$ after first applying the power rule of logarithms to the logarithmic expression on the right side of the equal sign. Once you have your formula for $\lz{y}{x}\text{,}$ substitute $x^x$ for $y\text{.}$ Voila! You will have the derivative formula for $x^x\text{.}$ So go ahead and do it.