Question

If \(e^y+xy = e\), then the ordered pair \((\frac{\text{d}y}{\text{d}x},\frac{\text{d}^2y}{\text{d}x^2})\) at \(x = 0\) is equal to

• A. \((\frac{1}{e},\frac{-1}{e^2})\)
• B. \((\frac{-1}{e},\frac{1}{e^2})\)
• C. \((\frac{1}{e},\frac{1}{e^2})\)
• D. \((\frac{-1}{e},\frac{-1}{e^2})\)

Hint:

Find y at x = 0, differentiate twice implicitly and substitute.

Answer 1

The Correct Option is B

Step 1: Point
$x = 0$ gives $y = 1$.

Step 2: First derivative
$y' = -\dfrac{y}{e^y + x}$. At $(0, 1)$ this is $-1/e$.

Step 3: Second derivative
Differentiating $e^y y' + y + x y' = 0$ again and substituting gives $e\,y'' = \frac{2}{e} - \frac{1}{e} = \frac1e$, so $y'' = 1/e^2$. Option (B).

Final Answer:
Option (B). \[ \boxed{\left(-\frac{1}{e},\frac{1}{e^{2}}\right)} \]