If y=tan(x+y), then dy/dx is equal to:

Options

A. `-cot^2(x+y)`
B. `-"cosec"^2(x+y)`
C. `-cot(x+y)"cosec"(x+y)`
D. `-tan^2(x+y)`

Answer

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

### Step 1: Differentiate both sides
We start with the equation:
\[ y = \tan(x + y) \]
Differentiating both sides with respect to \(x\):
\[ \frac{dy}{dx} = \sec^2(x + y) \cdot \left(1 + \frac{dy}{dx}\right) \]
This uses the chain rule, where the derivative of \(\tan(u)\) is \(\sec^2(u) \cdot \frac{du}{dx}\).

### Step 2: Expand the right-hand side
Now, we expand the right-hand side:
\[ \frac{dy}{dx} = \sec^2(x + y) + \sec^2(x + y) \cdot \frac{dy}{dx} \]

### Step 3: Collect all \(\frac{dy}{dx}\) terms
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} - \sec^2(x + y) \cdot \frac{dy}{dx} = \sec^2(x + y) \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left(1 - \sec^2(x + y)\right) = \sec^2(x + y) \]

### Step 4: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\sec^2(x + y)}{1 - \sec^2(x + y)} \]

### Step 5: Simplify the expression
Using the identity \(\sec^2(u) = 1 + \tan^2(u)\), we can rewrite the denominator:
\[ 1 - \sec^2(x + y) = 1 - (1 + \tan^2(x + y)) = -\tan^2(x + y) \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{\sec^2(x + y)}{-\tan^2(x + y)} = -\frac{\sec^2(x + y)}{\tan^2(x + y)} \]

### Final Answer
The final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -\cot^2(x + y) \]

---