Question

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

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

Hint:

Differentiate exponential functions: \(\frac{d}{dx}(a^x) = a^x \ln a\).

Answer 1

The Correct Option is C

To solve the given equation \(2^x + 2^y = 2^{x+y}\) for \(\frac{dy}{dx}\), we need to first find a relation between \(x\) and \(y\) and differentiate it with respect to \(x\). 

1. Since \(2^x + 2^y = 2^{x+y}\), we can manipulate the equation for simplification:

\[ 2^x + 2^y = 2^{x+y} \]

\[ \text{Divide throughout by } 2^x: \ 1 + 2^{y-x} = 2^y \]

1. Taking natural logarithms on both sides,

\[ \ln(1 + 2^{y-x}) = \ln(2^y) \]

1. Using the properties of logarithms,

\[ \ln(1 + 2^{y-x}) = y \ln(2) \]

1. Now differentiate both sides with respect to \(x\):

\[ \frac{d}{dx}[\ln(1 + 2^{y-x})] = \frac{d}{dx}[y \ln(2)] \]

\[ \frac{1}{1 + 2^{y-x}} \times \left(-2^{y-x} \ln(2) + 2^{y-x} \cdot \ln(2)\cdot \frac{dy}{dx}\right) = \ln(2) \cdot \frac{dy}{dx} \]

1. Equating and simplifying:

\[ \left(-2^{y-x} + 2^{y-x} \frac{dy}{dx}\right) = (1 + 2^{y-x}) \frac{dy}{dx} \]

\[ 2^{y-x} \frac{dy}{dx} - 2^{y-x} = \left(1 + 2^{y-x}\right) \frac{dy}{dx} \]

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

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

1. Switch sides and simplify:

\[ \frac{dy}{dx} = \frac{2^x(2^y - 1)}{2^y(1 - 2^x)} \]

Thus, the correct answer is \(\frac{2^x(2^y - 1)}{2^y(1 - 2^x)}\). The other options were either numerical manipulations or incorrect transformations that do not match with the differential equation derived from the original function.