Question

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

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. 1

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Simplify the trigonometric expression using double angle identities before differentiating. Recall that $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$.

Step 2: Detailed Explanation:
1. Simplify $f(x)$:
\[ f(x) = \frac{1 + \tan^2 x}{1 - \tan^2 x} = \frac{1}{\frac{1 - \tan^2 x}{1 + \tan^2 x}} = \frac{1}{\cos 2x} = \sec 2x \]
2. Differentiate $f(x) = \sec 2x$:
\[ f'(x) = 2 \sec 2x \tan 2x \]
3. Evaluate at $x = \frac{\pi}{8}$:
\[ f'(\frac{\pi}{8}) = 2 \sec(2 \cdot \frac{\pi}{8}) \tan(2 \cdot \frac{\pi}{8}) = 2 \sec(\frac{\pi}{4}) \tan(\frac{\pi}{4}) \]
4. Substitute values ($\sec(\pi/4) = \sqrt{2}$ and $\tan(\pi/4) = 1$):
\[ f'(\frac{\pi}{8}) = 2 \cdot \sqrt{2} \cdot 1 = 2\sqrt{2} \]

Step 3: Final Answer:
The value of $f'(\frac{\pi}{8})$ is $2\sqrt{2}$.