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Show that the derivative of an implicit function satisfies an equation

Consider the equation

    \[ x \sin (xy) + 2x^2 = 0. \]

Assuming the derivative y' exists, show that y satisfies the equation

    \[ y' x^2 \cos (xy) + (xy) \cos (xy)  + \sin (xy)+ 4x = 0. \]


We differentiate the given equation with respect to x (keeping in mind that y is a function of x, so we must use the chain rule).

    \begin{align*}  x \sin (xy) + 2x^2 = 0 && \implies && (x \sin (xy) + 2x^2)' &= 0 \\  && \implies && \sin (xy) + x (xy' + y) \cos (xy) + 4x &= 0 \\  && \implies && y' x^2 \cos (xy) + xy \cos (xy) + \sin (xy) + 4x &= 0. \end{align*}

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