How do you determine dy/dx given $x^{2} y + y = 3$ $x^2y+y=3$ ?

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Answer 1 · 2016-11-03T13:29:48

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

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

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

using the product rule on the first term and differentiating as normal for the rest.

$2 x y + x^{2} \frac{d y}{d x} + \frac{d y}{d x} = 0$ $2xy+x^2(dy)/(dx)+(dy)/(dx)=0$

rearranging for $\frac{d y}{d x}$ $(dy)/(dx)$

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

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

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

Answer 2 · 2016-11-03T14:06:32

Unless I was asked to use implicit differentiation, I would solve for $y$ $y$ first.

$x^{2} y + y = 3$ $x^2y+y=3$ is equivalent to

$y = \frac{3}{x^{2} + 1} = 3 {(x^{2} + 1)}^{- 1}$ $y = 3/(x^2+1) = 3(x^2+1)^-1$

So,

$\frac{d y}{d x} = - 3 {(x^{2} + 1)}^{- 2} \cdot [2 x]$ $dy/dx = -3(x^2+1)^-2 * [2x]$ $" "$ (Use the chain rule)

$= \frac{- 6 x}{{(x^{2} + 1)}^{2}}$ $= (-6x)/(x^2+1)^2$

How do you determine dy/dx given x^2y+y=3x2y+y=3x^2y+y=3?