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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

यदि \(e ^{ y }+ xy = e\), तो \(x =0\) पर क्रमित युग्म \(\left(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}\right)\) बराबर हैं 

  1. A \(\left( {\frac{1}{e}, - \frac{1}{{{e^2}}}} \right)\)
  2. B \(\left( {\frac{1}{e},  \frac{1}{{{e^2}}}} \right)\)
  3. C \(\left( { - \frac{1}{e},\frac{1}{{{e^2}}}} \right)\)
  4. D \(\left( { - \frac{1}{e}, - \frac{1}{{{e^2}}}} \right)\)
Verified Solution

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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