Question

If \[ 2^x+2^y=2^{x+y}, \] then \[ \frac{dy}{dx}= \]

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

Hint:

In implicit differentiation, remember that \(\frac{d}{dx}(a^y)=a^y\log a\cdot \frac{dy}{dx}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is an implicit differentiation problem involving exponential functions. The relationship between \( x \) and \( y \) is given by an equation that cannot be easily solved for \( y \) explicitly.
We will differentiate both sides with respect to \( x \) and solve for the derivative \( y' \).
Step 2: Key Formula or Approach:
1. Derivative of \( a^u \): \( \frac{d}{dx}(a^u) = a^u \ln a \frac{du}{dx} \).
2. Implicit differentiation principle.
Step 3: Detailed Explanation:

Differentiating the equation:
Differentiate \( 2^x + 2^y = 2^{x+y} \) with respect to \( x \):
\[ 2^x \ln 2 + 2^y \ln 2 \cdot \frac{dy}{dx} = 2^{x+y} \ln 2 \cdot \frac{d}{dx}(x+y) \]
\[ 2^x \ln 2 + 2^y \ln 2 \cdot \frac{dy}{dx} = 2^{x+y} \ln 2 \cdot \left(1 + \frac{dy}{dx}\right) \]

Simplifying:
Cancel the common factor \( \ln 2 \) from all terms:
\[ 2^x + 2^y \frac{dy}{dx} = 2^{x+y} + 2^{x+y} \frac{dy}{dx} \]

Isolating the derivative:
\[ 2^y \frac{dy}{dx} - 2^{x+y} \frac{dy}{dx} = 2^{x+y} - 2^x \]
\[ \frac{dy}{dx} (2^y - 2^{x+y}) = 2^{x+y} - 2^x \]
\[ \frac{dy}{dx} = \frac{2^{x+y} - 2^x}{2^y - 2^{x+y}} \]

Using the original equation to simplify:
Substitute \( 2^{x+y} = 2^x + 2^y \):
\[ \frac{dy}{dx} = \frac{(2^x + 2^y) - 2^x}{2^y - (2^x + 2^y)} \]
\[ \frac{dy}{dx} = \frac{2^y}{-2^x} = -2^{y-x} \]

Alternative simplification for options:
From \( 2^x + 2^y = 2^{x+y} \), divide by \( 2^{x+y} \): \( 2^{-y} + 2^{-x} = 1 \).
So, \( 2^{-x} = 1 - 2^{-y} \).
Then \( \frac{dy}{dx} = -2^{y-x} = -2^y \cdot 2^{-x} = -2^y (1 - 2^{-y}) = -2^y + 1 = 1 - 2^y \).

Step 4: Final Answer:
The derivative is \( 1 - 2^y \).