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

Answer

To find the derivative of \( y \) with respect to \( x \) for the equation \( \sin^2 y + \cos(xy) = \pi \), we will use implicit differentiation. Let's go through the steps:

### Step 1: Differentiate both sides of the equation
We start with the equation:
\[ \sin^2 y + \cos(xy) = \pi \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(\sin^2 y) + \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(\pi) \]
Since \( \pi \) is a constant, its derivative is \( 0 \).

### Step 2: Differentiate \( \sin^2 y \)
Using the chain rule:
\[ \frac{d}{dx}(\sin^2 y) = 2\sin y \cdot \frac{d}{dx}(\sin y) = 2\sin y \cdot \cos y \cdot \frac{dy}{dx} \]
So, we have:
\[ 2\sin y \cos y \cdot \frac{dy}{dx} \]

### Step 3: Differentiate \( \cos(xy) \)
Using the chain rule and product rule:
\[ \frac{d}{dx}(\cos(xy)) = -\sin(xy) \cdot \frac{d}{dx}(xy) \]
Now, applying the product rule to \( xy \):
\[ \frac{d}{dx}(xy) = x \cdot \frac{dy}{dx} + y \cdot \frac{d}{dx}(x) = x \cdot \frac{dy}{dx} + y \]
Thus, we have:
\[ -\sin(xy) \cdot (x \cdot \frac{dy}{dx} + y) \]

### Step 4: Combine the derivatives
Putting it all together, we have:
\[ 2\sin y \cos y \cdot \frac{dy}{dx} - \sin(xy)(x \cdot \frac{dy}{dx} + y) = 0 \]

### Step 5: Rearranging the equation
Now, we can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ 2\sin y \cos y \cdot \frac{dy}{dx} = \sin(xy)(x \cdot \frac{dy}{dx} + y) \]
Expanding the right side:
\[ 2\sin y \cos y \cdot \frac{dy}{dx} = \sin(xy) \cdot x \cdot \frac{dy}{dx} + \sin(xy) \cdot y \]

### Step 6: Factor out \( \frac{dy}{dx} \)
Now, we can factor \( \frac{dy}{dx} \) from both sides:
\[ (2\sin y \cos y - \sin(xy) \cdot x) \cdot \frac{dy}{dx} = \sin(xy) \cdot y \]

### Step 7: Solve for \( \frac{dy}{dx} \)
Finally, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\sin(xy) \cdot y}{2\sin y \cos y - \sin(xy) \cdot x} \]

### Final Answer
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{\sin(xy) \cdot y}{2\sin y \cos y - \sin(xy) \cdot x} \]