Question

If $2^x+2^y=2^{x+y}$, then the value of $\frac{d y}{d x}$ at $x=y=1$ is

(A) 0

(B) -1

(C) 1

(D) 2

Answer 1

On differentiating w.r.t. $x$, we get

$\begin{array}{l} 2^x(\log 2)+2^y(\log 2) \frac{d y}{d x}=2^{x+y}(\log 2)\left(1+\frac{d y}{d x}\right) \\\Rightarrow \quad 2^x+2^y \frac{d y}{d x}=2^{x+y}+2^{x+y}\left(\frac{d y}{d x}\right) \\\Rightarrow \quad \frac{d y}{d x}=\frac{2^{x+y}-2^x}{2^y-2^{x+y}}\end{array}$

Therefore, $\left(\frac{d y}{d x}\right)_{x=y=1}=\frac{2^2-2}{2-2^2}=\frac{2}{-2}=-1$

So, The correct option will be (B).