If x^2 e^ y , , + ,2xy e^ x , , + 13 , = ,0, then dy dx equals - Vidyadip

MCQ

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

✓
$- \frac{{2x{e^{y - x}} + 2y(x + 1)}}{{x(x{e^{y - x}} + 2)}}$
B
$ \frac{{2x{e^{x - y}} + 2y(x + 1)}}{{x(x{e^{y - x}} + 2)}}$
C
$- \frac{{2x{e^{x - y}} + 2y(x + 1)}}{{x(x{e^{x - y}} + 2)}}$
D
None of these

✓

Answer

Correct option: A.

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

a
Using partial derivatives, we have

$\frac{d y}{d x} =-\frac{2 x e^{y}+2 y e^{x}+2 x y e^{x}}{x^{2} e^{y}+2 x e^{x}} $

$=-\frac{2 x e^{y-x}+2 y+2 x y}{x^{2} e^{y-x}+2 x} $

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

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