Question

If \[ y = \log \left( x - \sqrt{x^2 - 1} \right), \] then \[ (x^2 - 1)y'' + xy' + e^y + \sqrt{x^2 - 1} = \] evaluates to:

• A. \( 0 \)
• B. \( 1 \)
• C. \( \sqrt{x^2 - 1} \)
• D. \( x \)

Hint:

For logarithmic differentiation, simplify expressions before differentiating to avoid complex fraction manipulations.

Answer 1

The Correct Option is D

Step 1: Differentiate \( y = \log (x - \sqrt{x^2 - 1}) \) 
Using the derivative of logarithm: \[ y' = \frac{1}{x - \sqrt{x^2 - 1}} \cdot \frac{d}{dx} (x - \sqrt{x^2 - 1}). \] Differentiating the expression inside: \[ \frac{d}{dx} (x - \sqrt{x^2 - 1}) = 1 - \frac{1}{2\sqrt{x^2 - 1}} \cdot \frac{d}{dx} (x^2 - 1). \] \[ = 1 - \frac{1}{2\sqrt{x^2 - 1}} \cdot (2x). \] \[ = 1 - \frac{x}{\sqrt{x^2 - 1}}. \] Thus, \[ y' = \frac{1 - \frac{x}{\sqrt{x^2 - 1}}}{x - \sqrt{x^2 - 1}}. \] Simplifying: \[ y' = \frac{\sqrt{x^2 - 1} - x}{(x - \sqrt{x^2 - 1})\sqrt{x^2 - 1}}. \] Since \( x - \sqrt{x^2 - 1} = \frac{1}{x + \sqrt{x^2 - 1}} \), \[ y' = \frac{1}{\sqrt{x^2 - 1}}. \] 

Step 2: Compute \( y'' \) 
Differentiating again: \[ y'' = -\frac{x}{(x^2 - 1)^{3/2}}. \] 

Step 3: Compute the given expression 
\[ (x^2 - 1)y'' + xy' + e^y + \sqrt{x^2 - 1}. \] Substituting values: \[ (x^2 - 1) \left( -\frac{x}{(x^2 - 1)^{3/2}} \right) + x \cdot \frac{1}{\sqrt{x^2 - 1}} + e^y + \sqrt{x^2 - 1}. \] \[ = -\frac{x}{\sqrt{x^2 - 1}} + \frac{x}{\sqrt{x^2 - 1}} + e^y + \sqrt{x^2 - 1}. \] \[ = 0 + e^y + \sqrt{x^2 - 1}. \] Since \( e^y = x - \sqrt{x^2 - 1} \), \[ = (x - \sqrt{x^2 - 1}) + \sqrt{x^2 - 1}. \] \[ = x. \]

 Step 4: Conclusion 
Thus, the correct answer is: \[ \mathbf{x}. \]