Question

If \[ \frac{x}{\sqrt{1 + x^2}} + \frac{y}{\sqrt{1 + y^2}} = 0, \quad x \neq y, \text{ then } (1 + x^2)^2 \frac{dy}{dx} = \]

• A. \( \frac{1}{2} \)
• B. 0
• C. -1
• D. 1

Hint:

When differentiating equations with both \( x \) and \( y \), use implicit differentiation and apply the chain rule carefully.

Answer 1

The Correct Option is C

Step 1: Differentiating the given equation.
We are given the equation \( \frac{x}{\sqrt{1 + x^2}} + \frac{y}{\sqrt{1 + y^2}} = 0 \). We differentiate both sides with respect to \( x \) implicitly: \[ \frac{d}{dx} \left( \frac{x}{\sqrt{1 + x^2}} \right) + \frac{d}{dx} \left( \frac{y}{\sqrt{1 + y^2}} \right) = 0 \]
Step 2: Applying the chain rule.
The derivative of \( \frac{x}{\sqrt{1 + x^2}} \) with respect to \( x \) is \( \frac{\sqrt{1 + x^2} - x \cdot \frac{x}{\sqrt{1 + x^2}}}{1 + x^2} \), and similarly for \( y \). Solving these, we get: \[ (1 + x^2)^2 \frac{dy}{dx} = -1 \]
Step 3: Conclusion.
Thus, the value of \( (1 + x^2)^2 \frac{dy}{dx} \) is -1, which makes option (C) the correct answer.