Question

If $2^x+2^y=2^{x+y}$, then $\frac{dy}{dx}$ has the value equal to

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

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $-\frac{2^y}{2^x}$
B $\frac{1}{1-2^x}$
C $1-2^y$
D $\frac{2^x(1-2^y)}{2^y(2^x-1)}$

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

Differentiating both the sides

$\Rightarrow \frac{d}{dx}(2^x+2^y)=\frac{d}{dx}\left(2^{x+y}\right)$

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

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

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

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

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