Discussion
If \(f(x) = \frac{\sqrt{2}\sin x}{\sqrt{1 + \cos 2x}}\), then \(f'\left(\frac{\pi}{6}\right) =\)
Answer: C
\(f'\left(\frac{\pi}{6}\right) =\) \(\frac{4}{3}\).
Since \(1 + \cos 2x = 2\cos^{2}x\), near \(\frac{\pi}{6}\) (where \(\cos x \gt 0\)):
\[ \begin{aligned} f(x) &= \frac{\sqrt{2}\sin x}{\sqrt{2}\cos x} \\ &= \tan x \\ f'(x) \\ &= \sec^{2}x \end{aligned} \]
so \(f'\left(\frac{\pi}{6}\right) = \frac{1}{\cos^{2}(\pi/6)}\) \(= \frac{1}{3/4} = \frac{4}{3}\). \(\frac{3}{4}\) is \(\cos^{2}\frac{\pi}{6}\) itself, and the other options do not follow from \(\sec^{2}x\).
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