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
- A \( \frac{1}{e}, \frac{1}{e^{2}} \)
- B \( -\frac{1}{e}, -\frac{1}{e^{2}} \)
- C \(-\frac{1}{e}, \frac{1}{e^{2}}\)
- D \( \frac{1}{e}, -\frac{1}{e^{2}} \)
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 \(x\): \(\mathrm{e}^{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{d} x} + \mathrm{y} + x \frac{\mathrm{dy}}{\mathrm{d} x} = 0\)
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