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

Concept: For implicit differentiation, differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\).

Step 1: Given: \[ 2^x+2^y=2^{x+y} \]

Step 2: Differentiate both sides. \[ 2^x\log 2+2^y\log 2\frac{dy}{dx} = 2^{x+y}\log 2\left(1+\frac{dy}{dx}\right) \] Cancel \(\log 2\): \[ 2^x+2^y\frac{dy}{dx} = 2^{x+y}\left(1+\frac{dy}{dx}\right) \]

Step 3: Since from the given equation: \[ 2^{x+y}=2^x+2^y \] So: \[ 2^x+2^y\frac{dy}{dx} = (2^x+2^y)\left(1+\frac{dy}{dx}\right) \]

Step 4: Expand right side. \[ 2^x+2^y\frac{dy}{dx} = 2^x+2^y+2^x\frac{dy}{dx}+2^y\frac{dy}{dx} \] Cancel common terms: \[ 0=2^y+2^x\frac{dy}{dx} \] \[ 2^x\frac{dy}{dx}=-2^y \] \[ \frac{dy}{dx}=-\frac{2^y}{2^x} \]

Step 5: From the original equation, divide by \(2^x\): \[ 1+2^{y-x}=2^y \] \[ 2^{y-x}=2^y-1 \] Thus: \[ \frac{dy}{dx}=-(2^y-1) \] \[ \frac{dy}{dx}=1-2^y \] Therefore, \[ \boxed{1-2^y} \]