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. $\frac{2^{x+y}-2^x}{2^y}$

Answer 1

Answer: (C)

Given $2^x+2^y=2^{x+y}$ On differentiating w.r.t. $x,$ we get $2^x{\log }_{e}2+2^y{\log }_e\frac{dy}{dx}$ $=2^{x+y}{\log }_{e}2\left(1+\frac{dy}{dx}\right)$ $\Rightarrow$ $\frac{dy}{dx}(2^y{\log }_{e}2-2^{x+y}{\log }_{e}2)$ $=2^{x+y}{\log }_{e}2(1-2^x)$ $\Rightarrow$ $\frac{dy}{dx}=[2^y{\log }_{e}2(1-2^x)]$ $=2^x{\log }_{e}2[2^y-1]$ $\Rightarrow$ $\frac{dy}{dx}=\frac{2^x}{2^y}\frac{2^y-1}{1-2^x}=2^{x-y}\left(\frac{2^y-1}{1-2^x}\right)$