Limits, Continuity & Differentiability
Differentiation of Logarithmic Functions
Grade 12
Question:
<p>If <span class="math">f(x) = \log_2(\log x)</span>, then <span class="math">f'(x)</span> at <span class="math">x = e</span> is</p>
<p>(a) 0</p>
<p>(b) 1</p>
<p>(c) <span class="math">e^{-1}</span></p>
<p>(d) <span class="math">(2e)^{-1}</span></p>
Step-by-Step Solution
Key Concept: Use chain rule for composite logarithmic functions and evaluate at x = e where log e = 1.
<p>We have, <span class="math">f(x) = \log_2(\log x) = \frac{\log(\log x)}{\log 2} = \frac{1}{2}\frac{\log(\log x)}{\log x}</span></p><p><span class="math">f'(x) = \frac{1}{2}\left[\frac{\log x \cdot \frac{1}{\log x} \cdot \frac{1}{x} - \log(\log x)}{(\log x)^2}\right]</span></p><p><span class="math">f'(e) = \frac{1}{2}\left[\frac{\frac{1}{e} - \log(\log e)}{(\log e)^2}\right] = \frac{1}{2} \cdot \frac{\frac{1}{e}}{1} = \frac{1}{2e} = (2e)^{-1}</span></p><p>∴ Answer is (d).</p>
Correct Answer: D