# How do you implicitly differentiate -1=y^3x-xy+x^4y ?

##### 1 Answer
Mar 23, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - {y}^{3} - 4 y {x}^{3}}{3 x {y}^{2} - x + {x}^{4}}$

#### Explanation:

The equation is:

${y}^{3} x - x y + {x}^{4} y = - 1$

Differentiating w.r.t. $x$

We have to apply product rule:

$\left[{y}^{3} \cdot \frac{d}{\mathrm{dx}} x + x \frac{d}{\mathrm{dx}} \cdot {y}^{3}\right] - \left[x \cdot \frac{d}{\mathrm{dx}} y + y \frac{d}{\mathrm{dx}} \cdot x\right] + \left[{x}^{4} \cdot \frac{d}{\mathrm{dx}} \cdot y + y \cdot \frac{d}{\mathrm{dx}} {x}^{4}\right] = 0$

$\left[{y}^{3} + 3 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right] - \left[x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y\right] + \left[{x}^{4} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + 4 y {x}^{3}\right] = 0$

${y}^{3} + 3 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - x \frac{\mathrm{dy}}{\mathrm{dx}} - y + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} + 4 y {x}^{3} = 0$

$3 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - x \frac{\mathrm{dy}}{\mathrm{dx}} + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{3} + y - 4 y {x}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[3 x {y}^{2} - x + {x}^{4}\right] = y - {y}^{3} - 4 y {x}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - {y}^{3} - 4 y {x}^{3}}{3 x {y}^{2} - x + {x}^{4}}$