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} \, log\, 2+2^{y} \,log\, 2\frac{dy}{dx} $
$=2^{\left(x+y\right)}\cdot log\,2 \left\{1+\frac{dy}{dx}\right\}$
$\Rightarrow 2^{x}+2^{y} \frac{dy}{dx}=2^{\left(x+y\right)}+2^{\left(x+y\right)} \frac{dy}{dx}$
$\Rightarrow \left(2^{x}-2^{x}\cdot2^{y}\right)=\left(2^{x}\cdot2^{y}-2^{y}\right)\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx} =-\frac{\left(2^{x}\cdot2^{y}-2^{x}\right)}{\left(2^{x}\cdot2^{y}-2y\right)}$
$\Rightarrow \left(\frac{dy}{dx}\right)_{at \left(1, 1\right)}$
$=-1$