The chain rule and logarithms, question 25: UBC MATH 100 practice

Question 25 of 30Challenging4 marks · 7 minTheory UBC Math 100 §4.4CLP-1 §2.10Key results

Let $f(x) = (2 + \cos x)^{x}$ . Find $f'\!\left(\dfrac{\pi}{2}\right)$ .

Your answer

An exact value, or a decimal to three places.

How sure are you?

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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 and chain rules, then multiply by f. Evaluate the trigonometric values first so everything stays exact.

Working

$\log f(x) = x\log(2 + \cos x)$
The base 2 + cos x is at least 1, so the logarithm is defined for every x.
$\frac{f'(x)}{f(x)} = \log(2 + \cos x) + x\cdot\frac{-\sin x}{2 + \cos x}$
Product rule on the right; the chain rule differentiates log(2 + cos x).
$\frac{f'(\pi/2)}{f(\pi/2)} = \log 2 - \frac{\pi/2}{2} = \log 2 - \frac{\pi}{4}, \qquad f\!\left(\tfrac{\pi}{2}\right) = 2^{\pi/2}$
Substitute cos(π/2) = 0 and sin(π/2) = 1.
$f'\!\left(\tfrac{\pi}{2}\right) = 2^{\pi/2}\left(\log 2 - \frac{\pi}{4}\right) \approx -0.274$
Multiply by f(π/2). The bracket is negative because π/4 ≈ 0.785 is larger than log 2 ≈ 0.693.

Answer

f'\!\left(\frac{\pi}{2}\right) = 2^{\pi/2}\left(\log 2 - \frac{\pi}{4}\right) \approx -0.274

Check

The falling base alone would give about −2.333 and the growing exponent alone about +2.059, so the net slope is small and negative, −0.274.

Where this goes wrong

Using the power rule as if the exponent were constant, x(2 + cos x)^x−1 · (−sin x), which gives about −2.333. When the base and the exponent both vary, take logarithms first.

The idea behind it

The chain rule and logarithms lesson, a figure to move, in a few steps

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 25