Question

Let \( y = \log_e \left( \frac{1 - x^2}{1 + x^2} \right), -1 < x < 1 \). Then at \( x = \frac{1}{2} \), the value of \( 225(y' - y'') \) is equal to

• A. 732
• B. 746
• C. 742
• D. 736

Answer 1

The Correct Option is D

Given: \(y = \log_e \left( \frac{1 - x^2}{1 + x^2} \right)\)

Step 1. First derivative \( \frac{dy}{dx} \):\(\frac{dy}{dx} = y' = \frac{-4x}{1 - x^4}\)

Step 2. Second derivative \( \frac{d^2y}{dx^2} \):\(y'' = \frac{-4(1 + 3x^4)}{(1 - x^4)^2}\)

Step 3. Calculate \( y' - y'' \) at \( x = \frac{1}{2} \): \(y' - y'' = \frac{-4x}{1 - x^4} + \frac{4(1 + 3x^4)}{(1 - x^4)^2}\)
 

Step 4. Substitute \( x = \frac{1}{2} \) and simplify to find \( 225(y' - y'') \): After calculations:   \(225(y' - y'') = 736\)
  
The Correct Answer is:736