If ` x^(3) +x^(2) y +xy^(2) +y^(3) =81,then (dy)/(dx) =`

Options

A. ` -((3x^(2)+ 2xy +y^(2))/( x^(2) +2xy +3y^(2)))`
B. ` -( x^(2) +2xy +3y^(2))/((3x^(2)+ 2xy +y^(2)))`
C. ` (3x^(2)+ 2xy +y^(2))/( x^(2) +2xy +3y^(2))`
D. `( x^(2) +2xy +3y^(2)) /(3x^(2)+ 2xy +y^(2))`

Answer

To solve the problem, we need to differentiate the given implicit function with respect to \( x \) and find \( \frac{dy}{dx} \).

Given:
\[ x^3 + x^2y + xy^2 + y^3 = 81 \]

### Step 1: Differentiate both sides with respect to \( x \)
Using implicit differentiation:
\[ \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) + \frac{d}{dx}(y^3) = \frac{d}{dx}(81) \]

### Step 2: Apply the differentiation rules
1. For \( x^3 \):
\[ \frac{d}{dx}(x^3) = 3x^2 \]

2. For \( x^2y \) (using the product rule):
\[ \frac{d}{dx}(x^2y) = x^2 \frac{dy}{dx} + 2xy \]

3. For \( xy^2 \) (using the product rule):
\[ \frac{d}{dx}(xy^2) = y^2 \frac{d}{dx}(x) + x \frac{d}{dx}(y^2) = y^2 + 2xy \frac{dy}{dx} \]

4. For \( y^3 \) (using the chain rule):
\[ \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} \]

5. The derivative of a constant (81) is:
\[ \frac{d}{dx}(81) = 0 \]

### Step 3: Combine all the differentiated terms
Putting it all together:
\[ 3x^2 + (x^2 \frac{dy}{dx} + 2xy) + (y^2 + 2xy \frac{dy}{dx}) + (3y^2 \frac{dy}{dx}) = 0 \]

### Step 4: Rearranging the equation
Combine like terms:
\[ 3x^2 + 2xy + y^2 + (x^2 + 2xy + 3y^2) \frac{dy}{dx} = 0 \]

### Step 5: Isolate \( \frac{dy}{dx} \)
Rearranging gives:
\[ (x^2 + 2xy + 3y^2) \frac{dy}{dx} = - (3x^2 + 2xy + y^2) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Finally, we can express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2} \]

### Final Answer
\[ \frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2} \]