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

Answer

To find the derivative of \( y \) with respect to \( x \) for the equation \( xy + y^2 = \tan x + y \), we will use implicit differentiation. Here’s the step-by-step solution:

### Step 1: Differentiate both sides of the equation
We start with the equation:
\[ xy + y^2 = \tan x + y \]
We will differentiate both sides with respect to \( x \).

### Step 2: Differentiate the left-hand side
Using the product rule on \( xy \) and the chain rule on \( y^2 \):
- The derivative of \( xy \) is \( x \frac{dy}{dx} + y \) (keeping \( y \) constant while differentiating \( x \)).
- The derivative of \( y^2 \) is \( 2y \frac{dy}{dx} \).

Thus, the left-hand side becomes:
\[ \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = x \frac{dy}{dx} + y + 2y \frac{dy}{dx} \]

### Step 3: Differentiate the right-hand side
The right-hand side is \( \tan x + y \):
- The derivative of \( \tan x \) is \( \sec^2 x \).
- The derivative of \( y \) is \( \frac{dy}{dx} \).

Thus, the right-hand side becomes:
\[ \frac{d}{dx}(\tan x) + \frac{d}{dx}(y) = \sec^2 x + \frac{dy}{dx} \]

### Step 4: Set the derivatives equal to each other
Now we equate the derivatives from both sides:
\[ x \frac{dy}{dx} + y + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx} \]

### Step 5: Rearrange the equation
We can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ (x + 2y) \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y \]
This simplifies to:
\[ (x + 2y - 1) \frac{dy}{dx} = \sec^2 x - y \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1} \]

### Final Answer
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1} \]