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 for \(\frac{dy}{dx}\), where the equation given is \(2^x + 2^y = 2^{x+y}\), we will use implicit differentiation. Let's differentiate both sides of the equation with respect to \(x\).

1. Differentiating \(2^x\) with respect to \(x\): \(\frac{d}{dx}(2^x) = 2^x \ln(2)\).
2. Differentiating \(2^y\) with respect to \(x\) involves using the chain rule: \(\frac{d}{dx}(2^y) = 2^y \ln(2) \frac{dy}{dx}\).
3. Differentiating the right side, \(2^{x+y}\), involves the chain rule as well: \(\frac{d}{dx}(2^{x+y}) = 2^{x+y} \ln(2) \left(1 + \frac{dy}{dx}\right)\).
4. Equating the derivatives:
   \(2^x \ln(2) + 2^y \ln(2) \frac{dy}{dx} = 2^{x+y} \ln(2) \left(1 + \frac{dy}{dx}\right)\).
5. Cancel out \(\ln(2)\) from both sides: \(2^x + 2^y \frac{dy}{dx} = 2^{x+y} \left(1 + \frac{dy}{dx}\right)\).
6. Expand the right side: \(2^x + 2^y \frac{dy}{dx} = 2^{x+y} + 2^{x+y} \frac{dy}{dx}\).
7. Rearranging terms to solve for \(\frac{dy}{dx}\):
   \(2^y \frac{dy}{dx} - 2^{x+y} \frac{dy}{dx} = 2^{x+y} - 2^x\).
8. Factoring out \(\frac{dy}{dx}\):
   \(\frac{dy}{dx} (2^y - 2^{x+y}) = 2^{x+y} - 2^x\).
9. Isolating \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{2^{x+y} - 2^x}{2^y - 2^{x+y}}\).
10. Simplify the expression for \(\frac{dy}{dx}\):
    \(\frac{dy}{dx} = \frac{2^x (2^y - 1)}{2^y (1 - 2^x)}\).

Therefore, the correct answer is \(\frac{2^x (2^y - 1)}{2^y (1 - 2^x)}\).