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 $C$.

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

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