If `xy^(2)= ax^(2) +bxy + y^(2)` then `(dy)/(dx)` =

Options

A. `(2ax +by -y^(2))/(2xy - bx -2y)`
B. `(2x^(2) + axy+y^(2))/(x^(2) +y^(2)+2xy)`
C. `(2ax + by+y^(2))/(2xy +bx -2y)`
D. `(ax- by -xy)/(xy +x^(2) -y^(2))`

Answer

To solve the equation \( xy^2 = ax^2 + bxy + y^2 \) for \( \frac{dy}{dx} \), we will differentiate both sides with respect to \( x \). Here’s the step-by-step solution:

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

### Step 2: Apply the product rule on the left side
For the left side \( xy^2 \), we apply the product rule:
\[ \frac{d}{dx}(xy^2) = x \frac{d}{dx}(y^2) + y^2 \frac{d}{dx}(x) \]
Using the chain rule on \( y^2 \):
\[ \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \]
Thus, we have:
\[ \frac{d}{dx}(xy^2) = x(2y \frac{dy}{dx}) + y^2(1) = 2xy \frac{dy}{dx} + y^2 \]

### Step 3: Differentiate the right side
Now we differentiate the right side \( ax^2 + bxy + y^2 \):
\[ \frac{d}{dx}(ax^2) + \frac{d}{dx}(bxy) + \frac{d}{dx}(y^2) \]
- For \( ax^2 \):
\[ \frac{d}{dx}(ax^2) = 2ax \]
- For \( bxy \) (using the product rule):
\[ \frac{d}{dx}(bxy) = b \left( x \frac{d}{dx}(y) + y \frac{d}{dx}(x) \right) = b \left( x \frac{dy}{dx} + y \right) \]
- For \( y^2 \):
\[ \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \]

Putting it all together, the right side becomes:
\[ 2ax + b\left(x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} \]

### Step 4: Set the derivatives equal
Now we equate the derivatives from both sides:
\[ 2xy \frac{dy}{dx} + y^2 = 2ax + b\left(x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 2xy \frac{dy}{dx} - b x \frac{dy}{dx} - 2y \frac{dy}{dx} = 2ax + by - y^2 \]
Factoring out \( \frac{dy}{dx} \):
\[ \left(2xy - bx - 2y\right) \frac{dy}{dx} = 2ax + by - y^2 \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Finally, we isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{2ax + by - y^2}{2xy - bx - 2y} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{2ax + by - y^2}{2xy - bx - 2y} \]