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
Updated On: Apr 20, 2024
- \(\left( \frac{1}{e}, \frac{1}{e^2} \right)\)
- \(\left( \frac{1}{e}, -\frac{1}{e^2} \right)\)
- \(\left( -\frac{1}{e}, -\frac{1}{e^2} \right)\)
- \(\left( -\frac{1}{e}, \frac{1}{e^2} \right)\)
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The Correct Option is D
Solution and Explanation
The correct answer is (D) :\(\left( -\frac{1}{e}, \frac{1}{e^2} \right)\).
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