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

Answer

To solve the problem where \( \cos(x+y) = y \sin x \) and we need to find \( \frac{dy}{dx} \), we can follow these steps:

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

We start with the equation:
\[ \cos(x+y) = y \sin x \]

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

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

Using the chain rule on the left side:
\[ -\sin(x+y) \cdot \left(1 + \frac{dy}{dx}\right) \]

### Step 3: Differentiate the right side using the product rule

Using the product rule on the right side:
\[ \frac{dy}{dx} \sin x + y \cos x \]

### Step 4: Set the derivatives equal to each other

Now we equate the derivatives from both sides:
\[ -\sin(x+y) \left(1 + \frac{dy}{dx}\right) = \frac{dy}{dx} \sin x + y \cos x \]

### Step 5: Expand and rearrange the equation

Expanding the left side gives:
\[ -\sin(x+y) - \sin(x+y) \frac{dy}{dx} = \frac{dy}{dx} \sin x + y \cos x \]

Rearranging terms to isolate \( \frac{dy}{dx} \):
\[ -\sin(x+y) = \frac{dy}{dx} \sin x + y \cos x + \sin(x+y) \frac{dy}{dx} \]

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

Factoring \( \frac{dy}{dx} \) out from the right side:
\[ -\sin(x+y) = \frac{dy}{dx} (\sin x + \sin(x+y)) + y \cos x \]

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

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

### Final Result

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