Question:
If \(e^y + xy = e\), the ordered pair \(\left(\frac{dy}{dx}, \frac{d^2y}{dx^2}\right)\) at \(x = 0\) is equal to :
If \(e^y + xy = e\), the ordered pair \(\left(\frac{dy}{dx}, \frac{d^2y}{dx^2}\right)\) at \(x = 0\) is equal to :
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At \(x=0\), \(e^y=e\) gives \(y=1\). Differentiate implicitly twice and substitute.
Updated On: Nov 5, 2026
- \(\left(\frac{1}{e}, -\frac{1}{e^2}\right)\)
- \(\left(-\frac{1}{e}, \frac{1}{e^2}\right)\)
- \(\left(-\frac{1}{e}, -\frac{1}{e^2}\right)\)
- \(\left(\frac{1}{e}, \frac{1}{e^2}\right)\)
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The Correct Option is B
Solution and Explanation
At \(x=0\): \(e^y=e\Rightarrow y=1\).
Differentiate: \(e^yy'+y+xy'=0\). At \(x=0\): \(e\,y'+1=0\Rightarrow y'=-\frac1e\).
Differentiate again: \(e^y(y')^2+e^yy''+y'+y'+xy''=0\). At \(x=0\): \(e\cdot\frac{1}{e^2}+e\,y''-\frac2e=0\Rightarrow\frac1e+e\,y''-\frac2e=0\Rightarrow y''=\frac{1}{e^2}\).
Ordered pair \(\left(-\frac1e,\frac1{e^2}\right)\).
Answer: (2) \(\left(-\frac{1}{e}, \frac{1}{e^2}\right)\)
Differentiate: \(e^yy'+y+xy'=0\). At \(x=0\): \(e\,y'+1=0\Rightarrow y'=-\frac1e\).
Differentiate again: \(e^y(y')^2+e^yy''+y'+y'+xy''=0\). At \(x=0\): \(e\cdot\frac{1}{e^2}+e\,y''-\frac2e=0\Rightarrow\frac1e+e\,y''-\frac2e=0\Rightarrow y''=\frac{1}{e^2}\).
Ordered pair \(\left(-\frac1e,\frac1{e^2}\right)\).
Answer: (2) \(\left(-\frac{1}{e}, \frac{1}{e^2}\right)\)
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