If `2^(x)+2^(y)=2^(x+y)` then `(dy)/(dx)`is equal to

Options

A. `-2^(y)/2^(x)`
B. `(1)/(1-2^(x))`
C. `1-2^(y)`
D. `(2^(x)(1-2^(y)))/(2^(y)(2^(x)-1))`

Answer

To solve the equation \(2^x + 2^y = 2^{x+y}\) and find \(\frac{dy}{dx}\), 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:
\[ \frac{d}{dx}(2^x) + \frac{d}{dx}(2^y) = \frac{d}{dx}(2^{x+y}) \]

Using the chain rule, we have:
\[ 2^x \ln(2) + 2^y \frac{dy}{dx} \ln(2) = 2^{x+y} \left(\ln(2) \cdot \frac{d}{dx}(x+y)\right) \]

Since \(\frac{d}{dx}(x+y) = 1 + \frac{dy}{dx}\), we can rewrite the right side:
\[ 2^x \ln(2) + 2^y \frac{dy}{dx} \ln(2) = 2^{x+y} \ln(2) (1 + \frac{dy}{dx}) \]

### Step 2: Simplify the equation

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

### Step 3: Rearranging terms

Rearranging the terms to isolate \(\frac{dy}{dx}\):
\[ 2^y \frac{dy}{dx} \ln(2) - 2^{x+y} \frac{dy}{dx} \ln(2) = 2^{x+y} \ln(2) - 2^x \ln(2) \]

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

### Step 4: Solve for \(\frac{dy}{dx}\)

Dividing both sides by \(\ln(2)\) (assuming \(\ln(2) \neq 0\)):
\[ \frac{dy}{dx} (2^y - 2^{x+y}) = 2^{x+y} - 2^x \]

Now, isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{2^{x+y} - 2^x}{2^y - 2^{x+y}} \]

### Step 5: Simplify further

We can simplify the expression:
\[ \frac{dy}{dx} = \frac{2^x (2^y - 1)}{2^y - 2^{x+y}} \]

This gives us the final expression for \(\frac{dy}{dx}\).

### Final Result
\[ \frac{dy}{dx} = \frac{2^x (2^y - 1)}{2^y (1 - 2^x)} \]