if `2^x+2^y=2^(x+y)` then the value of `(dy)/(dx)` at `x=y=1`

Options

A. 0
B. -1
C. 1
D. 2

Answer

To find the value of \(\frac{dy}{dx}\) at \(x = y = 1\) given the equation \(2^x + 2^y = 2^{x+y}\), we will follow these steps:

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

Starting with the equation:
\[ 2^x + 2^y = 2^{x+y} \]

We differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(2^x) + \frac{d}{dx}(2^y) = \frac{d}{dx}(2^{x+y}) \]

### Step 2: Apply the differentiation rules

Using the chain rule and the formula for the derivative of an exponential function \(a^u\) which is \(a^u \ln(a) \cdot \frac{du}{dx}\), we get:
\[ 2^x \ln(2) + 2^y \ln(2) \cdot \frac{dy}{dx} = 2^{x+y} \ln(2) \cdot \left(1 + \frac{dy}{dx}\right) \]

### Step 3: Rearranging the equation

Now, we can rearrange the equation:
\[ 2^x \ln(2) + 2^y \ln(2) \cdot \frac{dy}{dx} = 2^{x+y} \ln(2) + 2^{x+y} \ln(2) \cdot \frac{dy}{dx} \]

### Step 4: Collect terms involving \(\frac{dy}{dx}\)

Bringing all terms involving \(\frac{dy}{dx}\) to one side:
\[ 2^y \ln(2) \cdot \frac{dy}{dx} - 2^{x+y} \ln(2) \cdot \frac{dy}{dx} = 2^{x+y} \ln(2) - 2^x \ln(2) \]

Factoring out \(\frac{dy}{dx}\):
\[ \left(2^y \ln(2) - 2^{x+y} \ln(2)\right) \frac{dy}{dx} = 2^{x+y} \ln(2) - 2^x \ln(2) \]

### Step 5: Solve for \(\frac{dy}{dx}\)

Now we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{2^{x+y} \ln(2) - 2^x \ln(2)}{2^y \ln(2) - 2^{x+y} \ln(2)} \]

### Step 6: Substitute \(x = 1\) and \(y = 1\)

Substituting \(x = 1\) and \(y = 1\):
\[ \frac{dy}{dx} = \frac{2^{1+1} \ln(2) - 2^1 \ln(2)}{2^1 \ln(2) - 2^{1+1} \ln(2)} \]

This simplifies to:
\[ \frac{dy}{dx} = \frac{4 \ln(2) - 2 \ln(2)}{2 \ln(2) - 4 \ln(2)} = \frac{2 \ln(2)}{-2 \ln(2)} = -1 \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at \(x = y = 1\) is:
\[ \frac{dy}{dx} = -1 \]