If `tan(x+y)=e^(x+y)`, then `(dy)/(dx)`

Options

A. is always equal to -1
B. may or may not be equal to -1
C. `(dy)/(dx)` cannot be obtained
D. none of these

Explanation

To solve the problem where \( \tan(x+y) = e^{(x+y)} \) and we need to find \( \frac{dy}{dx} \), we will follow these steps:

### Step 1: Differentiate both sides
We start with the given equation:
\[ \tan(x+y) = e^{(x+y)} \]
Now, we differentiate both sides with respect to \( x \).

Using the chain rule on the left side:
\[ \frac{d}{dx}[\tan(x+y)] = \sec^2(x+y) \cdot \frac{d}{dx}(x+y) = \sec^2(x+y) \cdot (1 + \frac{dy}{dx}) \]

For the right side, we apply the chain rule as well:
\[ \frac{d}{dx}[e^{(x+y)}] = e^{(x+y)} \cdot \frac{d}{dx}(x+y) = e^{(x+y)} \cdot (1 + \frac{dy}{dx}) \]

### Step 2: Set the derivatives equal
Now we set the derivatives equal to each other:
\[ \sec^2(x+y) \cdot (1 + \frac{dy}{dx}) = e^{(x+y)} \cdot (1 + \frac{dy}{dx}) \]

### Step 3: Rearrange the equation
We can rearrange this equation to isolate \( \frac{dy}{dx} \):
\[ \sec^2(x+y) \cdot (1 + \frac{dy}{dx}) - e^{(x+y)} \cdot (1 + \frac{dy}{dx}) = 0 \]
Factoring out \( (1 + \frac{dy}{dx}) \):
\[ (1 + \frac{dy}{dx})(\sec^2(x+y) - e^{(x+y)}) = 0 \]

### Step 4: Solve for \( \frac{dy}{dx} \)
This gives us two cases:
1. \( 1 + \frac{dy}{dx} = 0 \) which leads to \( \frac{dy}{dx} = -1 \)
2. \( \sec^2(x+y) - e^{(x+y)} = 0 \) which we will analyze.

### Step 5: Analyze the second case
The second case \( \sec^2(x+y) = e^{(x+y)} \) implies that:
\[ 1 + \tan^2(x+y) = e^{(x+y)} \]
Since \( \tan(x+y) = e^{(x+y)} \), substituting gives:
\[ 1 + e^{2(x+y)} = e^{(x+y)} \]
This is a contradiction since \( e^{(x+y)} \) cannot equal \( 1 + e^{2(x+y)} \) for all \( x \) and \( y \).

### Conclusion
Thus, the only valid solution is from the first case:
\[ \frac{dy}{dx} = -1 \]