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: Understanding the Concept:
This is an implicit function. We differentiate both sides with respect to \( x \), treating \( y \) as a function of \( x \) and applying the chain rule[cite: 1].
Step 2: Key Formula or Approach:
The derivative of \( 2^u \) is \( 2^u \log 2 \cdot \frac{du}{dx} \)[cite: 1].
Step 3: Detailed Explanation:
Differentiate \( 2^x + 2^y = 2^{x+y} \): \[ 2^x \log 2 + 2^y \log 2 \frac{dy}{dx} = 2^{x+y} \log 2 \left(1 + \frac{dy}{dx}\right) \] Divide by \( \log 2 \): \[ 2^x + 2^y \frac{dy}{dx} = 2^{x+y} + 2^{x+y} \frac{dy}{dx} \] Rearrange to solve for \( \frac{dy}{dx} \): \[ 2^y \frac{dy}{dx} - 2^{x+y} \frac{dy}{dx} = 2^{x+y} - 2^x \] \[ \frac{dy}{dx} (2^y - 2^{x+y}) = 2^{x+y} - 2^x \] Using the original equation \( 2^{x+y} = 2^x + 2^y \): \[ \frac{dy}{dx} (2^y - (2^x + 2^y)) = (2^x + 2^y) - 2^x \] \[ \frac{dy}{dx} (-2^x) = 2^y \] \[ \frac{dy}{dx} = -\frac{2^y}{2^x} = -2^{y-x} \]
Step 4: Final Answer:
The derivative \( \frac{dy}{dx} \) is \( -2^{y-x} \).