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, $3^x+3^y=3^{x+y}\dots \left(i\right)$
Differentiating $\left(i\right)$ w.r.t. $x$, we get
$3^{x}log3+3^{y}log3\frac{dy}{dx}$
$=3^{x+y}log3\left(1+\frac{dy}{dx}\right)$
$\Rightarrow 3^x+3^y\frac{dy}{dx}$
$=3^{x+y}+3^{x+y}\frac{dy}{dx}$
$\Rightarrow \left(3^y-3^{x+y}\right)\frac{dy}{dx}$
$=3^{x+y}-3^x$
$\Rightarrow \frac{dy}{dx}=\frac{3^{x+y}-3^x}{3^y-3^{x+y}}$
$=\frac{3^x\left(3^y-1\right)}{3^y\left(1-3^x\right)}$
$\therefore {\left[\frac{dy}{dx}\right]}_{\left(1,1\right)}$
$=\frac{3\left(3-1\right)}{3\left(1-3\right)}$
$=\frac{2}{-2}$
$=-1$