Find the derivative of the following w.r.t. x :
`x^(3)+x^(2)y+xy^(2)+y^(3)=81`

Answer

To find the derivative of the equation \( x^3 + x^2y + xy^2 + y^3 = 81 \) with respect to \( x \), we will use implicit differentiation. Here are the steps to solve the problem:

### Step 1: Differentiate both sides of the equation
We start with the equation:
\[ x^3 + x^2y + xy^2 + y^3 = 81 \]

Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) + \frac{d}{dx}(y^3) = \frac{d}{dx}(81) \]

### Step 2: Apply the differentiation rules
1. The derivative of \( x^3 \) is:
\[ 3x^2 \]
2. For \( x^2y \), we use the product rule:
\[ \frac{d}{dx}(x^2y) = x^2 \frac{dy}{dx} + 2xy \]
3. For \( xy^2 \), we again use 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 \( y^3 \), we use the chain rule:
\[ \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} \]
5. The derivative of a constant (81) is:
\[ 0 \]

### Step 3: Combine all derivatives
Putting it all together, we have:
\[ 3x^2 + (x^2 \frac{dy}{dx} + 2xy) + (y^2 + 2xy \frac{dy}{dx}) + (3y^2 \frac{dy}{dx}) = 0 \]

### Step 4: Rearranging the equation
Now, we can rearrange the equation:
\[ 3x^2 + 2xy + y^2 + (x^2 + 2xy + 3y^2) \frac{dy}{dx} = 0 \]

### Step 5: Isolate \(\frac{dy}{dx}\)
We can isolate \(\frac{dy}{dx}\):
\[ (x^2 + 2xy + 3y^2) \frac{dy}{dx} = - (3x^2 + 2xy + y^2) \]
Thus,
\[ \frac{dy}{dx} = \frac{- (3x^2 + 2xy + y^2)}{(x^2 + 2xy + 3y^2)} \]

### Final Answer
The derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{- (3x^2 + 2xy + y^2)}{(x^2 + 2xy + 3y^2)} \]