Question

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

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

Answer 1

Answer: (C)

$2^{x - y}\left( \frac{2^{y} - 1}{1 - 2^{x}} \right)$

Answer 2

Here 2^x + 2^y = 2^{x + y}

Diff. w.r.t. x,

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

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

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

̃ $\frac{dy}{dx}$ = 2^{x – y} . $\left( \frac{2^{y} - 1}{1–2^{x}} \right)$.