Question

If \( y = \frac{1 + \tan^2 x}{1 - \tan^2 x} \), find \( y' \left( \frac{\pi}{8} \right) \), where \( 0<x<\frac{\pi}{4} \).

Hint:

When differentiating trigonometric functions, recognize standard identities such as \( \frac{1 + \tan^2 x}{1 - \tan^2 x} = \tan(2x) \), and use the chain rule for composite functions.

Answer 1

Step 1: Differentiate the given expression.
We are given the expression for \( y \): \[ y = \frac{1 + \tan^2 x}{1 - \tan^2 x} \] This is the formula for \( \tan(2x) \). Therefore, we can rewrite the expression as: \[ y = \tan(2x) \]
Step 2: Differentiate \( y \) with respect to \( x \).
Differentiating both sides with respect to \( x \): \[ y' = \frac{d}{dx} \left( \tan(2x) \right) \] Using the chain rule, we get: \[ y' = 2 \cdot \sec^2(2x) \]
Step 3: Evaluate \( y' \left( \frac{\pi}{8} \right) \).
Substitute \( x = \frac{\pi}{8} \) into the derivative: \[ y' \left( \frac{\pi}{8} \right) = 2 \cdot \sec^2\left(2 \cdot \frac{\pi}{8}\right) = 2 \cdot \sec^2\left(\frac{\pi}{4}\right) \] Since \( \sec\left( \frac{\pi}{4} \right) = \sqrt{2} \), we have: \[ y' \left( \frac{\pi}{8} \right) = 2 \cdot \left( \sqrt{2} \right)^2 = 2 \cdot 2 = 4 \]