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How to Find dy/dx, Implicit differentiation? 2xy^3 + y = 2x

1 Answer

Jun 25, 2018

$\frac{d y}{d x} = \frac{2 - 2 y^{3}}{6 x y^{2} + 1}$ $dy/dx= (2-2y^3)/(6xy^2+1)$

Explanation:

$\frac{d}{d x} (2 x y^{3} + y) = \frac{d}{d x} \cdot 2 x$ $d/dx(2xy^3+y)=d/dx*2x$
$2 \cdot \frac{d}{d x} (x y^{3}) + \frac{d}{d x} \cdot y = 2$ $2*d/dx(xy^3)+d/dx*y=2$
Use product rule on first differential

$2 \cdot (3 x y^{2} \cdot \frac{d y}{d x} + y^{3}) + \frac{d y}{d x} = 2$ $2*(3xy^2*dy/dx+y^3)+dy/dx=2$

Expand

$6 x y^{2} \cdot \frac{d y}{d x} + 2 y^{3} + \frac{d y}{d x} = 2$ $6xy^2*dy/dx+2y^3+dy/dx=2$

Collect like terms

$(6 x y^{2} + 1) \cdot \frac{d y}{d x} + 2 y^{3} = 2$ $(6xy^2+1)*dy/dx+2y^3=2$

Solve for $\frac{d y}{d x}$ $dy/dx$

$\frac{d y}{d x} = \frac{2 - 2 y^{3}}{6 x y^{2} + 1}$ $dy/dx=(2-2y^3)/(6xy^2+1)$

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