The chain rule and logarithms, question 23

Topic: The chain rule and logarithms
Difficulty: Challenging
Suggested time: 7 min
Marks: 4
Theory: UBC Math 100 §4.4 CLP-1 §2.10Key results

Let $f(x) = x^{\cos x}$ for $x > 0$ . Find $f'\!\left(\dfrac{\pi}{3}\right)$ .

Your answer

An exact expression, or a decimal to three decimal places.

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

Worked solution

Strategy

Take logarithms to bring the exponent down, differentiate with the product rule, then multiply by f. Leave log(π/3) as it is: it does not simplify, and it is small because π/3 is close to 1.

Working

$\log f(x) = \cos x\,\log x$
The log of a power brings the exponent down as a factor.
$\frac{f'(x)}{f(x)} = -\sin x\,\log x + \frac{\cos x}{x}$
Product rule on cos x times log x; the left side is (log f)′.
$\frac{f'(\pi/3)}{f(\pi/3)} = -\frac{\sqrt{3}}{2}\log\frac{\pi}{3} + \frac{1/2}{\pi/3} = \frac{3}{2\pi} - \frac{\sqrt{3}}{2}\log\frac{\pi}{3}$
Substitute sin(π/3) = √3/2 and cos(π/3) = 1/2.
$f\!\left(\tfrac{\pi}{3}\right) = \left(\tfrac{\pi}{3}\right)^{1/2} = \sqrt{\tfrac{\pi}{3}}$
The exponent at π/3 is cos(π/3) = 1/2.
$f'\!\left(\tfrac{\pi}{3}\right) = \sqrt{\tfrac{\pi}{3}}\left(\frac{3}{2\pi} - \frac{\sqrt{3}}{2}\log\frac{\pi}{3}\right) \approx 0.448$
Multiply by f(π/3).

Answer

f'\!\left(\frac{\pi}{3}\right) = \sqrt{\frac{\pi}{3}}\left(\frac{3}{2\pi} - \frac{\sqrt{3}}{2}\log\frac{\pi}{3}\right) \approx 0.448

Check

Freezing the exponent at ½ gives a slope of 1/(2√(π/3)) ≈ 0.489, and the falling exponent on a base just above 1 takes off about 0.041, leaving 0.448.

Where this goes wrong

Using the power rule as if the exponent were constant, cos x · x^{cos x − 1}, which gives about 0.489. With the variable in both the base and the exponent, take logarithms first.

Leans on

The product rule, from Differentiation rules

Try a similar question

Let

f(x) = \bigl(x^{2} + 1\bigr)^{x}

. Use logarithmic differentiation to find

f'(1)

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The chain rule and logarithms practice question 23