The chain rule and logarithms, question 9

Topic: The chain rule and logarithms
Difficulty: Introductory
Suggested time: 3 min
Marks: 2
Theory: UBC Math 100 §4.4 CLP-1 §2.10Key results

Logarithmic differentiation starts by differentiating $\log f(x)$ . Let $f$ be the function below, for $x > 0$ , where $\log$ is the natural logarithm. Find $\dfrac{d}{dx}\bigl(\log f(x)\bigr)$ .

f(x) = \frac{(x + 1)^{5}}{x^{2}\left(x^{2} + 3\right)^{3}}

Your answer

A formula in x. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.

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Theory UBC Math 100 §4.4 CLP-1 §2.10Key results

Worked solution

Strategy

Taking logarithms turns a quotient of powers into a sum of simple logarithms, one for each factor. Differentiate that sum term by term. The result is f′(x)/f(x); multiplying it by f(x) is the step that would give f′(x).

Working

$\begin{aligned} \log f(x) &= \log\Bigl((x + 1)^{5}\Bigr) - \log\left(x^{2}\right) \\ &\quad - \log\Bigl(\left(x^{2} + 3\right)^{3}\Bigr) \end{aligned}$
The quotient becomes a difference, and each factor of the denominator is subtracted. For x > 0 every factor is positive.
$= 5\log(x + 1) - 2\log x - 3\log\left(x^{2} + 3\right)$
Powers come down as multiples. The logarithm of the sum x² + 3 stays whole.
$\frac{d}{dx}\log f(x) = \frac{5}{x + 1} - \frac{2}{x} - 3\cdot\frac{2x}{x^{2} + 3}$
Each logarithm gives the derivative of its inside over the inside; the last one needs the inner derivative 2x.
$= \frac{5}{x + 1} - \frac{2}{x} - \frac{6x}{x^{2} + 3}$
Collect the constants. This is f′(x)/f(x).

Answer

\dfrac{d}{dx}\log f(x) = \dfrac{5}{x + 1} - \dfrac{2}{x} - \dfrac{6x}{x^{2} + 3}

Check

For large x, f(x) behaves like x⁵/x⁸ = x⁻³, whose log has derivative −3/x, and the answer behaves like 5/x − 2/x − 6/x = −3/x.

Where this goes wrong

Leaving the powers inside the logarithms, so that log((x + 1)⁵) becomes log(x + 1) with derivative 1/(x + 1). The log of a power is the power times the log: the 5, 2 and 3 come down as factors.