JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \({e^y} + xy = e\), the ordered pair \(\left( {\frac{{dy}}{{dx}},\frac{{{d^2}y}}{{d{x^2}}}} \right)\) at \(x = 0\) is equal to
- A \(\left( {\frac{1}{e}, - \frac{1}{{{e^2}}}} \right)\)
- B \(\left( {\frac{1}{e}, \frac{1}{{{e^2}}}} \right)\)
- C \(\left( { - \frac{1}{e},\frac{1}{{{e^2}}}} \right)\)
- D \(\left( { - \frac{1}{e}, - \frac{1}{{{e^2}}}} \right)\)
Answer & Solution
Correct Answer
(C) \(\left( { - \frac{1}{e},\frac{1}{{{e^2}}}} \right)\)
Step-by-step Solution
Detailed explanation
\({e^y} = xy = e\) differentiate w.r.t. \(x\) \({e^y}\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} + y = 0\) \({e^y}\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} + y = 0\)…
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