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

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

### Step 2: Simplify the equation

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

### Step 3: Rearrange the equation

Rearranging gives us:
\[ 2^y \ln(2) \frac{dy}{dx} - 2^{x+y} \ln(2) \frac{dy}{dx} = 2^{x+y} \ln(2) - 2^x \ln(2) \]

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

### Step 4: Factor out common terms

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

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

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

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

### Step 6: Simplify further

Factoring out \( 2^x \) from the numerator:
\[ \frac{dy}{dx} = \frac{2^x(2^y - 1)}{2^y - 2^{x+y}} \]

Factoring out \( 2^y \) from the denominator:
\[ \frac{dy}{dx} = \frac{2^x(2^y - 1)}{2^y(1 - 2^x)} \]

### Final Result

Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{2^x(2^y - 1)}{2^y(1 - 2^x)} \]