Find `dy/dx if 2x^2+3xy+by-11x+13y+c=0`

Answer

To find \(\frac{dy}{dx}\) for the equation \(2x^2 + 3xy + by - 11x + 13y + c = 0\), we will differentiate both sides of the equation with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate the equation**:
We start with the equation:
\[ 2x^2 + 3xy + by - 11x + 13y + c = 0 \]
We will differentiate each term with respect to \(x\).

2. **Differentiate each term**:
- The derivative of \(2x^2\) is \(4x\).
- For \(3xy\), we use the product rule. Let \(u = x\) and \(v = y\):
\[ \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \cdot 1 = x \frac{dy}{dx} + y \]
Therefore, the derivative of \(3xy\) is:
\[ 3(x \frac{dy}{dx} + y) = 3x \frac{dy}{dx} + 3y \]
- The derivative of \(by\) (where \(b\) is a constant) is:
\[ b \frac{dy}{dx} \]
- The derivative of \(-11x\) is \(-11\).
- The derivative of \(13y\) is:
\[ 13 \frac{dy}{dx} \]
- The derivative of \(c\) (a constant) is \(0\).

3. **Combine the derivatives**:
Putting it all together, we have:
\[ 4x + (3x \frac{dy}{dx} + 3y) + b \frac{dy}{dx} - 11 + 13 \frac{dy}{dx} = 0 \]
Simplifying this gives:
\[ 4x + 3x \frac{dy}{dx} + 3y + b \frac{dy}{dx} + 13 \frac{dy}{dx} - 11 = 0 \]

4. **Rearranging the equation**:
Combine the terms involving \(\frac{dy}{dx}\):
\[ (3x + b + 13) \frac{dy}{dx} + (4x + 3y - 11) = 0 \]

5. **Isolate \(\frac{dy}{dx}\)**:
Rearranging gives:
\[ (3x + b + 13) \frac{dy}{dx} = 11 - 3y - 4x \]
Therefore,
\[ \frac{dy}{dx} = \frac{11 - 3y - 4x}{3x + b + 13} \]

### Final Result:
\[ \frac{dy}{dx} = \frac{11 - 3y - 4x}{3x + b + 13} \]