Question

If $2^{x} + 2^{y} = 2^{x+y}$ then $\frac{dy}{dx} =$

• A. $1 - 2^{y}$
• B. $1 - \frac{1}{2^{y}}$
• C. $1 + 2^{x-y}$
• D. $-2^{y-x}$

Hint:

For $a^x + a^y = a^{x+y}$, the derivative is always $-a^{y-x}$.

Answer 1

The Correct Option is D

Step 1: Concept
Use implicit differentiation for equations where $y$ cannot be easily isolated.

Step 2: Meaning
Differentiate both sides with respect to $x$, remembering that $\frac{d}{dx}(2^y) = 2^y \log 2 \frac{dy}{dx}$.

Step 3: Analysis
$2^x \log 2 + 2^y \log 2 \frac{dy}{dx} = 2^{x+y} \log 2 (1 + \frac{dy}{dx})$. Canceling $\log 2$ and rearranging: $2^x + 2^y \frac{dy}{dx} = 2^{x+y} + 2^{x+y} \frac{dy}{dx} \implies \frac{dy}{dx}(2^y - 2^{x+y}) = 2^{x+y} - 2^x$.

Step 4: Conclusion
Since $2^{x+y} = 2^x + 2^y$, the equation simplifies to $\frac{dy}{dx}(-2^x) = 2^y$, which gives $\frac{dy}{dx} = -\frac{2^y}{2^x} = -2^{y-x}$.
Final Answer: (D)