`xsqrt(1+y)+ysqrt(1+x)=0`, then `(dy)/(dx)=`

Options

A. `1+x`
B. `(1+x)^-2`
C. `-(1+x)^-1`
D. `-(1+x)^(-2)`

Explanation

To solve the equation \( x\sqrt{1+y} + y\sqrt{1+x} = 0 \) and find \( \frac{dy}{dx} \), we will follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ x\sqrt{1+y} + y\sqrt{1+x} = 0 \]
Rearranging gives:
\[ x\sqrt{1+y} = -y\sqrt{1+x} \]

### Step 2: Squaring Both Sides
Square both sides to eliminate the square roots:
\[ (x\sqrt{1+y})^2 = (-y\sqrt{1+x})^2 \]
This simplifies to:
\[ x^2(1+y) = y^2(1+x) \]

### Step 3: Expanding Both Sides
Expanding both sides results in:
\[ x^2 + x^2y = y^2 + y^2x \]

### Step 4: Rearranging the Terms
Rearranging gives:
\[ x^2 - y^2 + x^2y - y^2x = 0 \]
Factoring out common terms:
\[ (x^2 - y^2) + xy(x - y) = 0 \]
Using the difference of squares:
\[ (x - y)(x + y) + xy(x - y) = 0 \]
Factoring out \( (x - y) \):
\[ (x - y)(x + y + xy) = 0 \]

### Step 5: Finding Critical Points
This gives us two cases:
1. \( x - y = 0 \) (which implies \( y = x \))
2. \( x + y + xy = 0 \)

### Step 6: Finding \( \frac{dy}{dx} \) for \( y = x \)
If \( y = x \), then:
\[ \frac{dy}{dx} = 1 \]

### Step 7: Finding \( \frac{dy}{dx} \) for \( x + y + xy = 0 \)
Now, we need to differentiate \( x + y + xy = 0 \) implicitly:
\[ 1 + \frac{dy}{dx} + y + x\frac{dy}{dx} = 0 \]
Rearranging gives:
\[ (1 + x)\frac{dy}{dx} + y + 1 = 0 \]
Solving for \( \frac{dy}{dx} \):
\[ (1 + x)\frac{dy}{dx} = - (y + 1) \]
\[ \frac{dy}{dx} = \frac{-(y + 1)}{1 + x} \]

### Step 8: Substituting \( y \)
From \( x + y + xy = 0 \), we can express \( y \) in terms of \( x \):
\[ y = -\frac{x}{1+x} \]
Substituting this into the derivative:
\[ \frac{dy}{dx} = \frac{-\left(-\frac{x}{1+x} + 1\right)}{1 + x} \]
Simplifying gives:
\[ \frac{dy}{dx} = \frac{1 - \frac{x}{1+x}}{1 + x} = \frac{\frac{1+x-x}{1+x}}{1+x} = \frac{1}{(1+x)^2} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -\frac{1}{1+x^2} \]