If `ycosx+xcosy=pi`, then `(dy)/(dx)` is equal to

Options

A. `(ysinx-cosy)/(cosx-xsiny)`
B. `(cosy-ysinx)/(xsiny+cosx)`
C. `(ysinx+cosy)/(cosx-xsiny)`
D. `(ysinx+cosy)/(cosx+xsiny)`

Answer

To find \(\frac{dy}{dx}\) for the equation \(y \cos x + x \cos y = \pi\), we will use implicit differentiation. Here are the steps:

### Step 1: Differentiate both sides of the equation
We start with the equation:
\[ y \cos x + x \cos y = \pi \]
Now, we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(y \cos x) + \frac{d}{dx}(x \cos y) = \frac{d}{dx}(\pi) \]

### Step 2: Apply the product rule
For the first term \(y \cos x\), we apply the product rule:
\[ \frac{d}{dx}(y \cos x) = y \frac{d}{dx}(\cos x) + \cos x \frac{dy}{dx} = -y \sin x + \cos x \frac{dy}{dx} \]
For the second term \(x \cos y\), we also apply the product rule:
\[ \frac{d}{dx}(x \cos y) = \cos y + x \frac{d}{dx}(\cos y) = \cos y - x \sin y \frac{dy}{dx} \]

### Step 3: Set the derivatives equal to zero
Since \(\frac{d}{dx}(\pi) = 0\), we can set up the equation:
\[ -y \sin x + \cos x \frac{dy}{dx} + \cos y - x \sin y \frac{dy}{dx} = 0 \]

### Step 4: Rearrange the equation
Now, we rearrange the equation to isolate \(\frac{dy}{dx}\):
\[ \cos x \frac{dy}{dx} - x \sin y \frac{dy}{dx} = y \sin x - \cos y \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}(\cos x - x \sin y) = y \sin x - \cos y \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{y \sin x - \cos y}{\cos x - x \sin y} \]

Thus, the final result is:
\[ \frac{dy}{dx} = \frac{y \sin x - \cos y}{\cos x - x \sin y} \]