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MHT CET · Maths · Differentiation

If \(\mathrm{e}^{\mathrm{y}}+x \mathrm{y}=\mathrm{e}\), then the ordered pair \(\left(\frac{\mathrm{dy}}{\mathrm{d} x}, \frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{d} x^2}\right)\) at \(x=0\) is equal to

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

Answer & Solution

Correct Answer

(C) \(-\frac{1}{e}, \frac{1}{e^{2}}\)

Step-by-step Solution

Detailed explanation

At \(x=0\): \(\mathrm{e}^{\mathrm{y}}+0 \cdot \mathrm{y}=\mathrm{e} \Rightarrow \mathrm{e}^{\mathrm{y}}=\mathrm{e} \Rightarrow \mathrm{y}=1\) Differentiating \(\mathrm{e}^{\mathrm{y}}+x \mathrm{y}=\mathrm{e}\) with respect to \(x\): \(\mathrm{e}^{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{d} x} + \mathrm{y} + x \frac{\mathrm{dy}}{\mathrm{d} x} = 0\)