Question

If $xy=e^{x-y}$ then

• A. $\frac{dy}{dx}$ doesn't exist at $x=0$
• B. $\frac{dy}{dx}=0$ when $x=1$
• C. $\frac{dy}{dx}=\frac{1}{2}$ when $x=0$
• D. none of these

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{dy}{dx}$ doesn't exist at $x=0$
B $\frac{dy}{dx}=0$ when $x=1$

$xy=e^{x-y}$
Differentiating with respect to x, we get
$y+x\frac{dy}{dx}=e^{x-y}(1-\frac{dy}{dx})$.
$y+x\frac{dy}{dx}=\left(xy\right)(1-\frac{dy}{dx})$.
$y+x\frac{dy}{dx}=xy-\frac{xy.dy}{dx}$
$\frac{dy}{dx}(x+xy)=xy-y$
$\frac{dy}{dx}=\frac{xy-y}{x+xy}$
Now
${\frac{dy}{dx}}_{|x=1}=\frac{y-y}{1+y}=0$.

Also, when $x=0$, we see that the denominator becomes $0$. So $\frac{dy}{dx}$ doesn't exist at $x=0$

Hence, option $A$ and $B$ are correct.