Question

If $ 3^{x} +3^{y} = 3^{x+y} $ , then at $ x = y = 1, \frac{dy}{dx} = $

• A. $ 1 $
• B. $ -1 $
• C. $ 0 $
• D. $ 2 $

Answer 1

Answer: (B)

We have, $3^{x}+3^{y}=3^{x+y} \ldots\left(i\right)$
Differentiating $\left(i\right)$ w.r.t. $x$, we get
$3^{x}\, log\,3+3^{y} log\,3 \frac{dy}{dx}$
$=3^{x+y} log\,3 \left(1+\frac{dy}{dx}\right)$
$\Rightarrow 3^{x}+3^{y} \frac{dy}{dx}$
$=3^{x+y}+3^{x+y} \frac{dy}{dx}$
$\Rightarrow \left(3^{y}-3^{x+y}\right)\frac{dy}{dx}$
$=3^{x+y}-3^{x}$
$\Rightarrow \frac{dy}{dx}=\frac{3^{x+y}-3^{x}}{3^{y}-3^{x+y}}$
$=\frac{3^{x}\left(3^{y}-1\right)}{3^{y}\left(1-3^{x}\right)}$
$\therefore \left[\frac{dy}{dx}\right]_{\left(1, 1\right)} $
$=\frac{3\left(3-1\right)}{3\left(1-3\right)}$
$=\frac{2}{-2}$
$=-1$