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How do you find the derivative of $y^{2} = 2 + x y$ $y^2=2+xy$ ?

1 Answer

Feb 24, 2017

$\frac{d y}{d x} = \frac{y}{2 y - x}$ $(dy)/(dx)=y/(2y-x)$

Explanation:

$y^{2} = 2 + x y$ $y^2=2+xy$

$\to XX y^{2} - x y = 2$ $rarr color(white)("XX")y^2-xy=2$

$\to XX \frac{d (y^{2} - x y)}{d x} = \frac{d (2)}{d x}$ $rarr color(white)("XX")(d(y^2-xy))/(dx)=(d (2))/(dx)$

$\to XX \frac{d (y^{2})}{d x} - \frac{d (x y)}{d x} = 0$ $rarr color(white)("XX")(d(y^2))/(dx)-(d(xy))/(dx)=0$

$\to XX 2 y \frac{d y}{d x} - (y + x \frac{d y}{d x}) = 0$ $rarr color(white)("XX")2y(dy)/(dx)-(y+x(dy)/(dx))=0$
$XXXXXXXXXXXX$ $color(white)("XXXXXXXXXXXX")$ (using the Chain and Product rules)

$\to XX 2 y \frac{d y}{d x} - x \frac{d y}{d x} = y$ $rarr color(white)("XX")2y(dy)/(dx)-x(dy)/(dx)=y$

$\to XX \frac{d y}{d x} (2 y - x) = y$ $rarr color(white)("XX")(dy)/(dx)(2y-x)=y$

$\to XX \frac{d y}{d x} = \frac{y}{2 y - x}$ $rarr color(white)("XX")(dy)/(dx)=y/(2y-x)$

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