Topic 10

Logarithmic Differentiation

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\log

3 kinds of question. Open the one you are stuck on.

Differentiate a log of a product or quotient by expanding first

\log\left(\dfrac{x^3\sqrt{x + 1}}{(x - 2)^2}\right)

1
$Split the log into a sum of simple logs$ $\log(ab) = \log a + \log b$
2
$\dfrac{d}{dx}\log u = \dfrac{u'}{u}$ $\log|u|$
3
$Simplify, or plug in the point$ $Combine over a common denominator only if the question asks for one form.$

Remember: Split first, then each log gives u' over u.

See it in a problem3 worked examples

f(x) = \log\left(\dfrac{x^3\sqrt{x + 1}}{(x - 2)^2}\right)

1
$Split the log into a sum of simple logs$ $f(x) = 3\log x + \tfrac{1}{2}\log(x + 1) - 2\log|x - 2|$
2
$\dfrac{d}{dx}\log u = \dfrac{u'}{u}$ $f'(x) = \dfrac{3}{x} + \dfrac{1}{2(x + 1)} - \dfrac{2}{x - 2}$
3
$Simplify, or plug in the point$ $Already simple: three fractions, no quotient rule. Leave it as a sum.$

f'(x) = \dfrac{3}{x} + \dfrac{1}{2(x + 1)} - \dfrac{2}{x - 2}

Common mistakes3

Splitting a sum: $\log(x^2 + 1)$ stays as it is. There is no rule for $\log(a + b)$ .
Dropping the coefficient: $\log\sqrt{x + 1} = \tfrac{1}{2}\log(x + 1)$ , so its derivative is $\dfrac{1}{2(x + 1)}$ .
Writing $(\log u)' = \dfrac{1}{u}$ and forgetting the $u'$ on top.

Differentiate a variable base raised to a variable power

y = x^{x}

1
$\log$ $y = f(x)^{g(x)}$
2
$\log y = g(x)\log f(x)$
3
$\dfrac{y'}{y}$ $The right side is a product, so use the product rule (and the chain rule inside the log).$
4
$y$

Remember: Log, drop the exponent, differentiate, multiply back.

See it in a problem4 worked examples

\dfrac{dy}{dx}

1
$\log$ $\log y = \log\big(x^{x}\big)$
2
$\log y = g(x)\log f(x)$ $\log y = x\log x$
3
$\dfrac{y'}{y}$ $\dfrac{y'}{y} = 1 \cdot \log x + x \cdot \dfrac{1}{x} = \log x + 1$
4
$y$ $y' = x^{x}(\log x + 1)$

\dfrac{dy}{dx} = x^{x}(\log x + 1)

Common mistakes4

Using the power rule: $(x^{x})' \neq x \cdot x^{x - 1}$ . The exponent is not a constant.
Using the exponential rule: $(x^{x})' \neq x^{x}\log x$ . The base is not a constant.
Stopping at $\dfrac{y'}{y} = \dots$ or leaving $y$ in the answer when it asks for "in terms of $x$ ".
Taking $\log$ of a sum such as $y = x^{x} + x^2$ . Log rules do not split sums: log-differentiate $x^{x}$ alone, then add $2x$ .

Use logarithmic differentiation on a messy product or quotient

\dfrac{(x^2 + 1)^3\sqrt{2x - 1}}{(x + 4)^5}

1
$\log$ $f$
2
$Expand with log rules until no products, quotients or powers remain$ $\log(ab) = \log a + \log b$
3
$\dfrac{f'}{f} = G(x)$
4
$f'(x) = f(x)\,G(x)$ $f'(a)$

Remember: Logs turn products into sums.

See it in a problem3 worked examples

f'(x)

1
$\log$ $\log f(x) = \log\left(\dfrac{(x^2 + 1)^3\sqrt{2x - 1}}{(x + 4)^5}\right)$
2
$Expand with log rules until no products, quotients or powers remain$ $\log f(x) = 3\log(x^2 + 1) + \tfrac{1}{2}\log(2x - 1) - 5\log(x + 4)$
3
$\dfrac{f'}{f} = G(x)$ $\dfrac{f'(x)}{f(x)} = \dfrac{6x}{x^2 + 1} + \dfrac{1}{2x - 1} - \dfrac{5}{x + 4}$
4
$f'(x) = f(x)\,G(x)$ $f'(x) = \dfrac{(x^2 + 1)^3\sqrt{2x - 1}}{(x + 4)^5}\left(\dfrac{6x}{x^2 + 1} + \dfrac{1}{2x - 1} - \dfrac{5}{x + 4}\right)$

f'(x) = f(x)\left(\dfrac{6x}{x^2 + 1} + \dfrac{1}{2x - 1} - \dfrac{5}{x + 4}\right)

Common mistakes3

"Expanding" $\log(a + b)$ as $\log a + \log b$ . Log rules work on products, never on sums.
Forgetting the chain rule inside each log: $\dfrac{d}{dx}\log(x^2 + 1) = \dfrac{2x}{x^2 + 1}$ .
Forgetting to multiply back by $f(x)$ at the end.

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Differentiate a log of a product or quotient by expanding first

Differentiate a variable base raised to a variable power

Use logarithmic differentiation on a messy product or quotient