Logarithmic Differentiation

Main idea

Logs turn multiplication into addition and pull exponents down in front. So when a function is a tangle of products, quotients and powers, take $\ln$ of both sides first, and the derivative becomes a sum of easy pieces. It's also how you handle functions like $x^x$ , where no other rule applies.

y = f(x) \implies {\htmlData{sym=0}{\frac{y'}{y}}} = \frac{d}{dx}\ln f(x)

What the symbols mean

$\frac{y'}{y}$
Relative rate of change
The derivative divided by the function: growth as a fraction of the current size.

Explanation

Logarithmic differentiation is simply a trick: take the log of a messy function first, because logs make it simpler, and then differentiate. It's good for two things: big products and quotients, and functions with $x$ in both the base and the exponent. The formula $\frac{y'}{y} = \frac{d}{dx}\ln f(x)$ looks odd, but it's just the chain rule on $\ln y$ : $\frac{y'}{y}$ (the derivative of $\ln y$ , which is $\frac1y$ times $y'$ ) equals the derivative of the log of your function. That ratio $\frac{y'}{y}$ even has a meaning of its own: the growth rate as a fraction of the current size.

This builds on two things you already have. Log rules turn products into sums, quotients into differences, and powers into multipliers. And implicit differentiation lets you differentiate $\ln y$ even though $y$ is the unknown function: the chain rule gives $\frac{1}{y}\cdot y'$ .

Steps:

Take $\ln$ of both sides and expand with log rules.
Differentiate implicitly: the left side becomes $\frac{y'}{y}$ .
Multiply through by $y$ and substitute the original function back in.

A common mistake. $x^x$ looks a lot like things you already know how to differentiate, so it's tempting to grab a familiar rule. The power rule would give $x\cdot x^{x-1}$ , and the exponential rule would give $x^x\ln x$ . Neither is right. The power rule needs a constant exponent, and the exponential rule needs a constant base, and $x^x$ has neither. Logs fix that: $\ln y = x\ln x$ is just a product, which you can handle. (Interestingly, the right answer turns out to be the sum of those two wrong ones.)

Big products and quotients. For something like $\frac{x^5(1 - 10x)}{\sqrt{x^2 + 2}}$ , the quotient rule with a product rule inside would be a mess. Taking logs turns it into three easy terms, one per factor.

See it

This is y = xˣ, with a tangent sliding along it. Logarithmic differentiation gives its slope as xˣ(ln x + 1). As you can see, it dips first and then climbs steeply: the tangent is flat at x = 1/e ≈ 0.37, where ln x = −1, and that's the lowest point, at a height of about 0.69. By x = 2 the slope is nearly 7.

Worked examples

Example 1

Differentiate $y = x^x$ .

1
Take $\ln$ of both sides, and the exponent comes down: $\ln y = x\ln x$ .

Example 2

Differentiate $y = \dfrac{x^5(1 - 10x)}{\sqrt{x^2 + 2}}$ .

1
Take $\ln$ and split it up with log rules. The square root is a power of $\frac12$ : $\ln y = 5\ln x + \ln(1 - 10x) - \frac12\ln(x^2 + 2)$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the slope of $y = x^x$ at $x = 1$ .

Problem 2+5 XP

Differentiate $x^x$ .

Problem 3+5 XP

Differentiate $x^{\sin x}$ .

Problem 4+5 XP

For $y = \dfrac{x^2e^x}{x + 1}$ , find $\dfrac{y'}{y}$ .

Problem 5+5 XP

Differentiate $(x^2 + 1)^3(x - 2)^2$ .

Problem 6+5 XP

A country's GDP grows $4\%$ per year and its population grows $1.5\%$ per year. How fast does GDP per person grow, in percent per year?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Differentiate $y = (\sin x)^{\ln x}$ for $0 < x < \pi$ .

Where it's used

Relative rates $\frac{y'}{y}$ are everywhere: growth rates in economics, percentage error propagation in physics (the error in $x^2y^3$ is about $2\frac{\Delta x}{x} + 3\frac{\Delta y}{y}$ ).

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