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
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
Show Hint
Find y at x = 0, differentiate twice implicitly and substitute.
Updated On: Oct 1, 2026
- \((\frac{1}{e},\frac{-1}{e^2})\)
- \((\frac{-1}{e},\frac{1}{e^2})\)
- \((\frac{1}{e},\frac{1}{e^2})\)
- \((\frac{-1}{e},\frac{-1}{e^2})\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Differentiate \(e^y + xy = e\) implicitly twice, then evaluate at \(x = 0\).
Step 2: Find y
At \(x = 0\): \(e^y = e\), so \(y = 1\).
Step 3: First derivative
\[ e^y y' + y + x y' = 0 \]
At \(x = 0\), \(y = 1\): \(e\,y' + 1 = 0\), so \(y' = -\frac1e\).
Step 4: Second derivative
Differentiate again: \(e^y (y')^2 + e^y y'' + y' + y' + x y'' = 0\).
At \(x = 0\): \(e \cdot \frac{1}{e^2} + e\,y'' - \frac2e = 0 \Rightarrow \frac1e + e\,y'' - \frac2e = 0\).
\[ y'' = \frac{1}{e^2} \]
So the pair is \(\left(-\frac1e, \frac{1}{e^2}\right)\), option (B).
Final Answer:
The ordered pair is \(\left(-\frac1e, \frac1{e^2}\right)\), option (B). \[ \boxed{\left(-\frac{1}{e},\frac{1}{e^{2}}\right)} \]
Differentiate \(e^y + xy = e\) implicitly twice, then evaluate at \(x = 0\).
Step 2: Find y
At \(x = 0\): \(e^y = e\), so \(y = 1\).
Step 3: First derivative
\[ e^y y' + y + x y' = 0 \]
At \(x = 0\), \(y = 1\): \(e\,y' + 1 = 0\), so \(y' = -\frac1e\).
Step 4: Second derivative
Differentiate again: \(e^y (y')^2 + e^y y'' + y' + y' + x y'' = 0\).
At \(x = 0\): \(e \cdot \frac{1}{e^2} + e\,y'' - \frac2e = 0 \Rightarrow \frac1e + e\,y'' - \frac2e = 0\).
\[ y'' = \frac{1}{e^2} \]
So the pair is \(\left(-\frac1e, \frac{1}{e^2}\right)\), option (B).
Final Answer:
The ordered pair is \(\left(-\frac1e, \frac1{e^2}\right)\), option (B). \[ \boxed{\left(-\frac{1}{e},\frac{1}{e^{2}}\right)} \]
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