Question

Encontre
dy/dx
derivando implicitamente:
$xy + {x}^{2} {y}^{3} = \sin(xy) + { {ye}^{x} }^{2}$
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Answer 1

$\mathsf{xy + {x}^{2} {y}^{3} = sen(xy) + { {ye}^{x} }^{2}}$

$\mathsf{y+x\dfrac{dy}{dx}+2xy^3+x^2.3y^2\dfrac{dy}{dx}=cos(xy)(y+x\dfrac{dy}{dx})+\dfrac{dy}{dx}e^{x^2}+y.e^{x^2}.2x}$

$\mathsf{y+x\dfrac{dy}{dx}+2xy^3+3x^2y^2\dfrac{dy}{dx} =ycos(xy)+xcos(xy)\dfrac{dy}{dx}+\dfrac{dy}{dx}e^{x^2}+2xye^{x^2}}$

$\mathsf{x\dfrac{dy}{dx}+3x^2y^2\dfrac{dy}{dx} -xcos(xy)\dfrac{dy}{dx} -\dfrac{dy}{dx}e^{x^2}=ycos(xy)+3xye^{x^2}-y-2xy^3}$

$\mathsf{\dfrac{dy}{dx}(x+3x^2y^2-xcos(xy)-e^{x^2})=ycos(xy) +3xye^{x^2}-y-2xy^3}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{ycos(xy)+3xye^{x^2}-y-2xy^3}{x+3x^2y^2-xcos(xy)-e^{x^2}}}$