Logarithmic differentiation

In $y=x^x$ $(x>0)$, $x$ appears in both the base and the exponent, so neither the formula for $(x^n)'$ nor the one for $(a^x)'$ applies directly. Take the logarithm of both sides instead ($\log_e$ in the examples is the natural logarithm $\log$).

By the chain rule, the left side $\log y$ differentiates to $\frac{y'}{y}$. Finally, multiply both sides by $y$ and substitute the original expression for $y$.

Examples

1. 01What is y'?$\displaystyle \frac{y'}{y} = \log_e x + 1$AnswerHide

   Answer$\displaystyle y(\log_e x + 1)$

   Steps

   1. Just multiply both sides by $y$
   2. That $y$ is the original function; at the end you write it back, as in $y=x^{x}$
2. 02Which one has $x$ in both the base and the exponent, so the formulas cannot be applied directly?$\displaystyle$A$\displaystyle y=e^{x}$B$\displaystyle y=2^{x}$C$\displaystyle y=x^{2}$D$\displaystyle y=x^{x}$AnswerHide

   Answer$\displaystyle y=x^{x}$

   Steps

   1. Only $y=x^{x}$: with $x$ as both base and exponent, neither the power rule ($x^{n}$) nor the exponential rule ($a^{x}$) fits
   2. $e^{x}$ and $2^{x}$ have constant bases, so the exponential rule handles them; $x^{2}$ has a constant exponent, so the power rule does
3. 03For a function y=f(x) that takes positive values, we took the logarithm of both sides and differentiated. What is □?$\displaystyle (\log_e y)' = \square$AnswerHide

   Answer$\displaystyle \frac{y'}{y}$

   Steps

   1. $\log_e y$ is a composition through $y$: the outer $\log$ differentiates to $\frac{1}{y}$ and the inner $y$ to $y'$
   2. Multiplying gives $\frac{y'}{y}$; the $y'$ appears because $y$ depends on $x$
4. 04What is the derivative of the right-hand side?$\displaystyle \log_e y = x\log_e x$AnswerHide

   Answer$\displaystyle \log_e x + 1$

   Steps

   1. Product rule: $x$ differentiates to 1 and $\log_e x$ to $\frac{1}{x}$
   2. $\displaystyle 1 \cdot \log_e x + x \cdot \frac{1}{x}$$\displaystyle = \log_e x + 1$
   3. In the second term the $x$ cancels to 1; that is where the $+1$ comes from
5. 05Taking the logarithm of both sides, what is $\log_e y$?$\displaystyle y=x^{3x}$AnswerHide

   Answer$\displaystyle 3x\log_e x$

   Steps

   1. An exponent comes down in front of the $\log$: $\log_e x^{3x} = 3x\log_e x$
   2. It does not become a sum: $x^{3x}$ is $3x$ copies of $x$ multiplied, so its $\log$ is $3x$ times as big
6. 06Taking the logarithm of both sides, what is $\log_e y$?$\displaystyle y=x^{x}$AnswerHide

   Answer$\displaystyle x\log_e x$

   Steps

   1. An exponent comes down in front of the $\log$: $\log_e x^{x} = x\log_e x$
   2. It does not become a sum: $x^{x}$ is $x$ copies of $x$ multiplied, so its $\log$ is $x$ times as big

FAQ

Q1What should I watch for when taking logarithms?AnswerHide

A

The argument must be positive. $y=x^x$ is considered for $x>0$, so $y>0$ and you can take $\log y$ directly. For a function that can take negative values, take $\log|y|$.

Q2Can I stop at $\frac{y'}{y}=\log x+1$?AnswerHide

A

Not yet. Multiply both sides by $y$ to get $y'=y(\log x+1)$, then substitute $y=x^x$ to finish with $y'=x^x(\log x+1)$.