# How do you use implicit differentiation to find dy/dx given xy^2+x^2y=x?

Nov 5, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 x y - {y}^{2}}{2 x y + {x}^{2}}$

#### Explanation:

The differentiation of the given expression is determined by using differentiation of the sum and the product differentiation.

Differentiation of the sum:

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(u + v\right) = \frac{\mathrm{du}}{\mathrm{dx}} + \frac{\mathrm{dv}}{\mathrm{dx}}}$

Product differentiation:

$\textcolor{b r o w n}{\left(u v\right) ' = u ' v + v ' u}$

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

$\Rightarrow \frac{d}{\mathrm{dx}} \left(x {y}^{2} + {x}^{2} y\right) = \frac{\mathrm{dx}}{\mathrm{dx}}$

$\Rightarrow \textcolor{b l u e}{\frac{d \left(x {y}^{2}\right)}{\mathrm{dx}} + \frac{d \left({x}^{2} y\right)}{\mathrm{dx}}} = 1$

rArrcolor(brown)((y^2xxdx/dx+x xx(dy^2)/dx))+color(brown)((yxxdx^2/dx+x^2xxdy/dx)=1

$\Rightarrow {y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\Rightarrow 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y + {y}^{2} = 1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(2 x y + {x}^{2}\right) + 2 x y + {y}^{2} = 1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(2 x y + {x}^{2}\right) = 1 - 2 x y - {y}^{2}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 x y - {y}^{2}}{2 x y + {x}^{2}}$