Question

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

• A. 0
• B. -1
• C. 1
• D. 2

Answer 1

Answer: (B)

$2^x+2^y=2^{x+y}$
On differentiating w. r.t x,
$2^{x}log2+2^{y}log2\frac{dy}{dx}$
$=2^{\left(x+y\right)}\cdot log2\left\{1+\frac{dy}{dx}\right\}$
$\Rightarrow 2^x+2^y\frac{dy}{dx}=2^{\left(x+y\right)}+2^{\left(x+y\right)}\frac{dy}{dx}$
$\Rightarrow \left(2^x-2^x\cdot 2^y\right)=\left(2^x\cdot 2^y-2^y\right)\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=-\frac{\left(2^x\cdot 2^y-2^x\right)}{\left(2^x\cdot 2^y-2y\right)}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{at\left(1,1\right)}$
$=-1$