Q. 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) $\frac{2^{x + y} - 2^{x}}{2^{y}}$
Explanation
(c)
On differentiating w.r.t. $x,\$we get
$2^{x}\log{2 + 2^{y}\log{2\frac{dy}{dx} = 2^{x + y}\log{2\left( 1 + \frac{dy}{dx} \right)}}}$
$\Rightarrow 2^{x} + 2^{y}\frac{dy}{dx} = \ 2^{x + y}\left( 1 + \frac{dy}{dx} \right)$
$\Rightarrow \ 2^{x} - 2^{x + y} = \frac{dy}{dx}\left( 2^{x + y} - 2^{y} \right)$
$\Rightarrow {\ 2}^{x - y}\frac{(1 - 2^{y})}{(2^{x} - 1)} = \frac{dy}{dx}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.