Obtain `dy/dx` when :
`x^(2)+y^(2)+2axy = 0`

Answer

To find \(\frac{dy}{dx}\) for the equation \(x^2 + y^2 + 2axy = 0\), we will use implicit differentiation. Here’s a step-by-step solution:

### Step 1: Differentiate both sides of the equation
We start with the equation:
\[ x^2 + y^2 + 2axy = 0 \]
Now, we differentiate both sides with respect to \(x\).

### Step 2: Apply the differentiation
Differentiating each term:
- The derivative of \(x^2\) is \(2x\).
- The derivative of \(y^2\) is \(2y \frac{dy}{dx}\) (using the chain rule).
- For the term \(2axy\), we use the product rule:
\[ \frac{d}{dx}(2axy) = 2a \left( x \frac{dy}{dx} + y \right) \]
So, the differentiation gives us:
\[ 2x + 2y \frac{dy}{dx} + 2a \left( x \frac{dy}{dx} + y \right) = 0 \]

### Step 3: Combine like terms
Now, we can rearrange the equation:
\[ 2x + 2y \frac{dy}{dx} + 2axy \frac{dy}{dx} + 2ay = 0 \]
Combining the terms involving \(\frac{dy}{dx}\):
\[ 2x + 2ay + (2y + 2ax) \frac{dy}{dx} = 0 \]

### Step 4: Isolate \(\frac{dy}{dx}\)
Now, we isolate \(\frac{dy}{dx}\):
\[ (2y + 2ax) \frac{dy}{dx} = -2x - 2ay \]
Dividing both sides by \(2y + 2ax\):
\[ \frac{dy}{dx} = \frac{-2x - 2ay}{2y + 2ax} \]

### Step 5: Simplify the expression
We can simplify this expression:
\[ \frac{dy}{dx} = \frac{-2(x + ay)}{2(y + ax)} = \frac{-(x + ay)}{y + ax} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{-(x + ay)}{y + ax} \]