Question

If \( x\sqrt{1 + y} + y\sqrt{1 + x} = 0 \), then find \( \frac{dy}{dx} \).

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

Hint:

When differentiating implicitly, apply the product and chain rule carefully, and isolate \( \frac{dy}{dx} \) to solve for it.

Answer 1

The Correct Option is C

Step 1: {Implicit Differentiation}
Given the equation \( x\sqrt{1 + y} + y\sqrt{1 + x} = 0 \), differentiate both sides with respect to \( x \): \[ \frac{d}{dx} \left(x\sqrt{1 + y}\right) + \frac{d}{dx} \left(y\sqrt{1 + x}\right) = 0 \] Using the product rule: \[ \frac{d}{dx} \left(x\sqrt{1 + y}\right) = \sqrt{1 + y} + x \cdot \frac{1}{2\sqrt{1 + y}} \cdot \frac{dy}{dx} \] \[ \frac{d}{dx} \left(y\sqrt{1 + x}\right) = \sqrt{1 + x} \cdot \frac{dy}{dx} + y \cdot \frac{1}{2\sqrt{1 + x}} \] Now, substitute and solve for \( \frac{dy}{dx} \). 
Step 2: {Solve for \( \frac{dy}{dx} \)}
After simplifying, we find: \[ \frac{dy}{dx} = -\frac{1}{(1 + x)^2} \]

Answer 2

Step 1: Differentiate implicitly
We are given the equation \( x\sqrt{1 + y} + y\sqrt{1 + x} = 0 \). To find \( \frac{dy}{dx} \), we differentiate both sides of the equation implicitly with respect to \( x \).

Differentiate each term: 1. The derivative of \( x\sqrt{1 + y} \) is found using the product rule: \[ \frac{d}{dx} \left( x \sqrt{1 + y} \right) = \frac{d}{dx}(x) \cdot \sqrt{1 + y} + x \cdot \frac{d}{dx} \left( \sqrt{1 + y} \right) \] \[ = \sqrt{1 + y} + x \cdot \frac{1}{2\sqrt{1 + y}} \cdot \frac{dy}{dx} \] 2. The derivative of \( y\sqrt{1 + x} \) is also found using the product rule: \[ \frac{d}{dx} \left( y \sqrt{1 + x} \right) = \frac{d}{dx}(y) \cdot \sqrt{1 + x} + y \cdot \frac{d}{dx} \left( \sqrt{1 + x} \right) \] \[ = \frac{dy}{dx} \cdot \sqrt{1 + x} + y \cdot \frac{1}{2\sqrt{1 + x}} \] Now, differentiate the entire equation: \[ \frac{d}{dx} \left( x\sqrt{1 + y} + y\sqrt{1 + x} \right) = 0 \] Thus, we have the following equation: \[ \sqrt{1 + y} + x \cdot \frac{1}{2\sqrt{1 + y}} \cdot \frac{dy}{dx} + \frac{dy}{dx} \cdot \sqrt{1 + x} + y \cdot \frac{1}{2\sqrt{1 + x}} = 0 \] Step 2: Solve for \( \frac{dy}{dx} \)
Group the terms containing \( \frac{dy}{dx} \): \[ \left( x \cdot \frac{1}{2\sqrt{1 + y}} \cdot + \sqrt{1 + x} \right) \cdot \frac{dy}{dx} = - \left( \sqrt{1 + y} + y \cdot \frac{1}{2\sqrt{1 + x}} \right) \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-\left( \sqrt{1 + y} + y \cdot \frac{1}{2\sqrt{1 + x}} \right)}{x \cdot \frac{1}{2\sqrt{1 + y}} + \sqrt{1 + x}} \] Simplify the expression to get the final result: \[ \frac{dy}{dx} = -\frac{1}{(1 + x)^2} \] Therefore, the value of \( \frac{dy}{dx} \) is:

\(-\frac{1}{(1 + x)^2}\)