Differentiate `sec(x+y) = xy`

Explanation

To differentiate the equation \( \sec(x+y) = xy \) with respect to \( x \), we will follow these steps:

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

We start with the equation:
\[ \sec(x+y) = xy \]

Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}[\sec(x+y)] = \frac{d}{dx}[xy] \]

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

Using the chain rule for differentiation on the left side:
\[ \frac{d}{dx}[\sec(x+y)] = \sec(x+y) \tan(x+y) \cdot \frac{d}{dx}(x+y) \]
The derivative of \( x+y \) with respect to \( x \) is \( 1 + \frac{dy}{dx} \):
\[ \sec(x+y) \tan(x+y) \cdot (1 + \frac{dy}{dx}) \]

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

Now, we differentiate the right side \( xy \) using the product rule:
\[ \frac{d}{dx}[xy] = x \frac{dy}{dx} + y \cdot 1 = x \frac{dy}{dx} + y \]

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

Now we have:
\[ \sec(x+y) \tan(x+y) \cdot (1 + \frac{dy}{dx}) = x \frac{dy}{dx} + y \]

### Step 5: Rearrange the equation to isolate \( \frac{dy}{dx} \)

Distributing on the left side:
\[ \sec(x+y) \tan(x+y) + \sec(x+y) \tan(x+y) \frac{dy}{dx} = x \frac{dy}{dx} + y \]

Now, we can rearrange this to isolate \( \frac{dy}{dx} \):
\[ \sec(x+y) \tan(x+y) \frac{dy}{dx} - x \frac{dy}{dx} = y - \sec(x+y) \tan(x+y) \]

Factoring out \( \frac{dy}{dx} \):
\[ \left( \sec(x+y) \tan(x+y) - x \right) \frac{dy}{dx} = y - \sec(x+y) \tan(x+y) \]

### Step 6: Solve for \( \frac{dy}{dx} \)

Finally, we solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{y - \sec(x+y) \tan(x+y)}{\sec(x+y) \tan(x+y) - x} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{y - \sec(x+y) \tan(x+y)}{\sec(x+y) \tan(x+y) - x} \]