Question:medium
Let $f(x) = \frac{1 + \tan^2 x}{1 - \tan^2 x}$ for $0 < x < \frac{\pi}{4}$. Then the value of $f'(\frac{\pi}{8})$ is equal to
Let $f(x) = \frac{1 + \tan^2 x}{1 - \tan^2 x}$ for $0 < x < \frac{\pi}{4}$. Then the value of $f'(\frac{\pi}{8})$ is equal to
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Always try to simplify a complex-looking trigonometric function into a single basic function. Differentiating $\sec 2x$ is far easier than using the quotient rule on the original expression.
Updated On: Jun 24, 2026
- $\frac{1}{\sqrt{2}}$
- $\sqrt{2}$
- $2\sqrt{2}$
- 1
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The Correct Option is C
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