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

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(A) Given the implicit equation $F(x, y) = {x^2}{e^y} + 2xy{e^x} + 13 = 0$.Using the formula for the derivative of an implicit function,$\frac{dy}{dx}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the implicit equation $F(x, y) = {x^2}{e^y} + 2xy{e^x} + 13 = 0$.Using the formula for the derivative of an implicit function,$\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$.First,find the partial derivative with respect to $x$:$\frac{\partial F}{\partial x} = 2x{e^y} + 2y{e^x} + 2xy{e^x}$.Next,find the partial derivative with respect to $y$:$\frac{\partial F}{\partial y} = {x^2}{e^y} + 2x{e^x}$.Now,substitute these into the formula:$\frac{dy}{dx} = -\frac{2x{e^y} + 2y{e^x} + 2xy{e^x}}{{x^2}{e^y} + 2x{e^x}}$.Divide the numerator and denominator by ${e^x}$:$\frac{dy}{dx} = -\frac{2x{e^{y-x}} + 2y + 2xy}{{x^2}{e^{y-x}} + 2x}$.Factor out $2$ in the numerator and $x$ in the denominator:$\frac{dy}{dx} = -\frac{2(x{e^{y-x}} + y(1+x))}{x(x{e^{y-x}} + 2)}$.This matches option $A$.

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

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