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

Answer

To find \(\frac{dy}{dx}\) for the equation \( \cos(xy) = \sin(x+y) \), we will use implicit differentiation. Here’s the step-by-step solution:

### Step 1: Differentiate both sides with respect to \(x\)

Starting with the equation:
\[ \cos(xy) = \sin(x+y) \]

We differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}[\cos(xy)] = \frac{d}{dx}[\sin(x+y)] \]

### Step 2: Apply the chain rule on the left side

Using the chain rule on the left side:
\[ -\sin(xy) \cdot \frac{d}{dx}(xy) \]

For \(\frac{d}{dx}(xy)\), we use the product rule:
\[ \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \]

So, the left side becomes:
\[ -\sin(xy) \cdot (x \frac{dy}{dx} + y) \]

### Step 3: Differentiate the right side

Now, differentiate the right side:
\[ \frac{d}{dx}[\sin(x+y)] = \cos(x+y) \cdot \frac{d}{dx}(x+y) \]

And since \(\frac{d}{dx}(x+y) = 1 + \frac{dy}{dx}\), we have:
\[ \cos(x+y) \cdot (1 + \frac{dy}{dx}) \]

### Step 4: Set the derivatives equal

Now we set the derivatives equal:
\[ -\sin(xy) \cdot (x \frac{dy}{dx} + y) = \cos(x+y) \cdot (1 + \frac{dy}{dx}) \]

### Step 5: Expand and rearrange

Expanding both sides gives:
\[ -\sin(xy) \cdot x \frac{dy}{dx} - \sin(xy) \cdot y = \cos(x+y) + \cos(x+y) \cdot \frac{dy}{dx} \]

Now, we want to collect all terms involving \(\frac{dy}{dx}\) on one side:
\[ -\sin(xy) \cdot x \frac{dy}{dx} - \cos(x+y) \cdot \frac{dy}{dx} = \cos(x+y) + \sin(xy) \cdot y \]

### Step 6: Factor out \(\frac{dy}{dx}\)

Factoring out \(\frac{dy}{dx}\) gives:
\[ \frac{dy}{dx} \left(-\sin(xy) \cdot x - \cos(x+y)\right) = \cos(x+y) + \sin(xy) \cdot y \]

### Step 7: Solve for \(\frac{dy}{dx}\)

Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\cos(x+y) + \sin(xy) \cdot y}{-\sin(xy) \cdot x - \cos(x+y)} \]

### Final Answer

Thus, the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\cos(x+y) + y \sin(xy)}{-x \sin(xy) - \cos(x+y)} \]