If ` cos (xy) =x+ y ,then (dy)/(dx)=`

Options

A. `-( (1+ xsin (xy))/(1+ysin (xy)))`
B. `-((1+ysin (xy))/ (1+ xsin (xy) ) ) `
C. `(1+ysin (xy))/ (1+ xsin (xy) ) `
D. ` (1+ xsin (xy) ) /(1+ysin (xy))`

Answer

To solve the problem where \( \cos(xy) = x + y \) and we need to find \( \frac{dy}{dx} \), we will use implicit differentiation. Here’s the step-by-step solution:

### Step 1: Differentiate both sides with respect to \( x \)
We start by differentiating the equation \( \cos(xy) = x + y \) with respect to \( x \).

\[ \frac{d}{dx}[\cos(xy)] = \frac{d}{dx}[x + y] \]

### Step 2: Apply the chain rule on the left side
Using the chain rule on the left side, we have:

\[ -\sin(xy) \cdot \frac{d}{dx}[xy] = 1 + \frac{dy}{dx} \]

### Step 3: Differentiate \( xy \) using the product rule
Now, we need to differentiate \( xy \) using the product rule:

\[ \frac{d}{dx}[xy] = x \frac{dy}{dx} + y \]

Substituting this back into the equation gives:

\[ -\sin(xy) \cdot (x \frac{dy}{dx} + y) = 1 + \frac{dy}{dx} \]

### Step 4: Distribute \( -\sin(xy) \)
Distributing \( -\sin(xy) \) results in:

\[ -\sin(xy) \cdot x \frac{dy}{dx} - y \sin(xy) = 1 + \frac{dy}{dx} \]

### Step 5: Rearrange the equation
Now, we rearrange the equation to isolate terms involving \( \frac{dy}{dx} \):

\[ -\sin(xy) \cdot x \frac{dy}{dx} - \frac{dy}{dx} = 1 + y \sin(xy) \]

Factoring out \( \frac{dy}{dx} \):

\[ \frac{dy}{dx}(-\sin(xy) \cdot x - 1) = 1 + y \sin(xy) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{1 + y \sin(xy)}{-\sin(xy) \cdot x - 1} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{1 + y \sin(xy)}{-\sin(xy) \cdot x - 1} \]

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