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:
We must find the derivative of the given trigonometric function and evaluate it at a specific angle.
Simplifying the function using trigonometric identities before differentiating makes it significantly easier.
Step 2: Key Formula or Approach:
Recall the double angle identity for cosine: \(\cos(2x) = \frac{1 - \tan^2 x}{1 + \tan^2 x}\).
The given function is the reciprocal of this identity.
Differentiate \(f(x) = \sec(2x)\) using the chain rule.
Step 3: Detailed Explanation:
Simplify \(f(x)\):
\[ f(x) = \frac{1 + \tan^2 x}{1 - \tan^2 x} = \frac{1}{\frac{1 - \tan^2 x}{1 + \tan^2 x}} = \frac{1}{\cos(2x)} = \sec(2x) \] Now, find the derivative \(f'(x)\):
\[ f'(x) = \frac{d}{dx}(\sec(2x)) = \sec(2x)\tan(2x) \cdot \frac{d}{dx}(2x) \] \[ f'(x) = 2\sec(2x)\tan(2x) \] Evaluate the derivative at \(x = \frac{\pi}{8}\):
\[ f'\left(\frac{\pi}{8}\right) = 2\sec\left(2 \cdot \frac{\pi}{8}\right)\tan\left(2 \cdot \frac{\pi}{8}\right) \] \[ f'\left(\frac{\pi}{8}\right) = 2\sec\left(\frac{\pi}{4}\right)\tan\left(\frac{\pi}{4}\right) \] Substitute standard trigonometric values (\(\sec(\pi/4) = \sqrt{2}\) and \(\tan(\pi/4) = 1\)):
\[ f'\left(\frac{\pi}{8}\right) = 2(\sqrt{2})(1) = 2\sqrt{2} \] Step 4: Final Answer:
The value is \(2\sqrt{2}\).