If ${x^2}{e^y} + 2xy{e^x} + 13 = 0,$ then $\frac{{dy}}{{dx}}$ equals

$\frac{{2x{e^{x - y}} + 2y(x + 1)}}{{x(x{e^{y - x}} + 2)}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

{x^2}{e^y} + 2xy{e^x} + 13 = 0,

then

\frac{{dy}}{{dx}}

equals

View answer and explanation

Correct option: B

- \left[ {\frac{{2x{e^{y - x}} + 2y(x + 1)}}{{x(x{e^{y - x}} + 2)}}} \right]

Option B (

- \left[ {\frac{{2x{e^{y - x}} + 2y(x + 1)}}{{x(x{e^{y - x}} + 2)}}} \right]

\) is correct because it represents the derivative dy/dx obtained through implicit differentiation of the given equation.

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