Question

If 2^x + 2^y = 2^x+y, then \(\frac{dy}{dx}\) is

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

Answer 1

The Correct Option is B

If \(2^x + 2^y = 2^{x+y}\), then we need to find \(\frac{dy}{dx}\).

Differentiate both sides with respect to x using implicit differentiation:

\(\frac{d}{dx}(2^x) + \frac{d}{dx}(2^y) = \frac{d}{dx}(2^{x+y})\)

Using the chain rule and the derivative of \(a^x\) is \(a^x \ln a\):

\(2^x \ln 2 + 2^y \ln 2 \frac{dy}{dx} = 2^{x+y} \ln 2 (1 + \frac{dy}{dx})\)

Divide by \(\ln 2\):

\(2^x + 2^y \frac{dy}{dx} = 2^{x+y} (1 + \frac{dy}{dx})\)

\(2^x + 2^y \frac{dy}{dx} = 2^{x+y} + 2^{x+y} \frac{dy}{dx}\)

Isolate \(\frac{dy}{dx}\):

\(2^y \frac{dy}{dx} - 2^{x+y} \frac{dy}{dx} = 2^{x+y} - 2^x\)

\(\frac{dy}{dx} (2^y - 2^{x+y}) = 2^{x+y} - 2^x\)

\(\frac{dy}{dx} = \frac{2^{x+y} - 2^x}{2^y - 2^{x+y}}\)

Since \(2^x + 2^y = 2^{x+y}\), substitute \(2^{x+y}\) in the above expression:

\(\frac{dy}{dx} = \frac{(2^x + 2^y) - 2^x}{2^y - (2^x + 2^y)}\)

\(\frac{dy}{dx} = \frac{2^y}{ -2^x}\)

\(\frac{dy}{dx} = -2^{y-x}\)

Therefore, the correct option is (B) \(-2^{y-x}\).

Answer 2

Given $ 2^x + 2^y = 2^{x+y} $, we want to find $ \frac{dy}{dx} $.

Differentiating with respect to $ x $, we have:

$$ 2^x \ln 2 + 2^y \ln 2 \frac{dy}{dx} = 2^{x+y} \ln 2 (1 + \frac{dy}{dx}) $$

Dividing by $ \ln 2 $:

$$ 2^x + 2^y \frac{dy}{dx} = 2^{x+y} (1 + \frac{dy}{dx}) $$

Simplify:

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

Rearrange terms:

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

Factorize:

$$ (2^y - 2^{x+y}) \frac{dy}{dx} = 2^{x+y} - 2^x $$

Solve for $ \frac{dy}{dx} $:

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

Since $ 2^{x+y} = 2^x + 2^y $, substitute this into the equation:

$$ \frac{dy}{dx} = \frac{2^x + 2^y - 2^x}{2^y - (2^x + 2^y)} $$

Simplify:

$$ \frac{dy}{dx} = \frac{2^y}{-2^x} = -2^{y-x} $$