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

Nov 3, 2016

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

#### Explanation:

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

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

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

$2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

rearranging for $\frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

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

Nov 3, 2016

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

#### Explanation:

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

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

So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {\left({x}^{2} + 1\right)}^{-} 2 \cdot \left[2 x\right]$ $\text{ }$ (Use the chain rule)

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