If {x^2}{e^y} + 2xy{e^x} + 13 = 0,then frac{d | vedclass

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Question

(C) Given the implicit function $f(x, y) = x^2 e^y + 2xy e^x + 13 = 0$. To find $\frac{dy}{dx}$,we use the formula $\frac{dy}{dx} = - \frac{\frac{\par

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the implicit function $f(x, y) = x^2 e^y + 2xy e^x + 13 = 0$. To find $\frac{dy}{dx}$,we use the formula $\frac{dy}{dx} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$. First,calculate the partial derivative with respect to $x$: $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 e^y + 2xy e^x + 13) = 2x e^y + 2y e^x + 2xy e^x = 2x e^y + 2y e^x(1 + x)$. Next,calculate the partial derivative with respect to $y$: $\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 e^y + 2xy e^x + 13) = x^2 e^y + 2x e^x$. Now,substitute these into the formula: $\frac{dy}{dx} = - \frac{2x e^y + 2y e^x(1 + x)}{x^2 e^y + 2x e^x}$. Divide the numerator and the denominator by $x e^x$: $\frac{dy}{dx} = - \frac{\frac{2x e^y}{e^x} + 2y(1 + x)}{\frac{x^2 e^y}{e^x} + 2x} = - \frac{2x e^{y-x} + 2y(x + 1)}{x(x e^{y-x} + 2)}$. Thus,the correct option is $C$.

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{-y(2x + y)}{x(x + 2y)}$

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

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