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

Options

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

Answer

To solve the equation \( 2^x + 2^y = 2^{x+y} \) for \( \frac{dy}{dx} \), we will differentiate both sides with respect to \( x \).

### Step-by-Step Solution:

1. **Differentiate Both Sides:**
Start with the equation:
\[ 2^x + 2^y = 2^{x+y} \]
Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(2^x) + \frac{d}{dx}(2^y) = \frac{d}{dx}(2^{x+y}) \]

2. **Apply the Differentiation Formula:**
Using the formula \( \frac{d}{dx}(a^u) = a^u \ln(a) \cdot \frac{du}{dx} \):
- For \( 2^x \):
\[ \frac{d}{dx}(2^x) = 2^x \ln(2) \]
- For \( 2^y \):
\[ \frac{d}{dx}(2^y) = 2^y \ln(2) \cdot \frac{dy}{dx} \]
- For \( 2^{x+y} \):
\[ \frac{d}{dx}(2^{x+y}) = 2^{x+y} \ln(2) \cdot (1 + \frac{dy}{dx}) \]

3. **Substitute the Derivatives:**
Substitute these derivatives back into the differentiated equation:
\[ 2^x \ln(2) + 2^y \ln(2) \cdot \frac{dy}{dx} = 2^{x+y} \ln(2) (1 + \frac{dy}{dx}) \]

4. **Simplify the Equation:**
We can divide the entire equation by \( \ln(2) \) (assuming \( \ln(2) \neq 0 \)):
\[ 2^x + 2^y \frac{dy}{dx} = 2^{x+y} (1 + \frac{dy}{dx}) \]

5. **Rearrange the Equation:**
Rearranging gives:
\[ 2^x + 2^y \frac{dy}{dx} = 2^{x+y} + 2^{x+y} \frac{dy}{dx} \]
Now, isolate the terms involving \( \frac{dy}{dx} \):
\[ 2^y \frac{dy}{dx} - 2^{x+y} \frac{dy}{dx} = 2^{x+y} - 2^x \]

6. **Factor Out \( \frac{dy}{dx} \):**
Factor \( \frac{dy}{dx} \) out from the left side:
\[ \frac{dy}{dx} (2^y - 2^{x+y}) = 2^{x+y} - 2^x \]

7. **Solve for \( \frac{dy}{dx} \):**
Finally, divide both sides by \( (2^y - 2^{x+y}) \):
\[ \frac{dy}{dx} = \frac{2^{x+y} - 2^x}{2^y - 2^{x+y}} \]

8. **Simplify the Expression:**
Notice that \( 2^{x+y} = 2^x \cdot 2^y \):
\[ \frac{dy}{dx} = \frac{2^x(2^y - 1)}{2^y(1 - 2^x)} \]

### Final Result:
Thus, the value of \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{2^y - 1}{2^y(1 - 2^x)} \]