# How do you implicitly differentiate -y= x^4y-3x^2y^2+xy^4 ?

Sep 17, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 {x}^{3} y - 6 x {y}^{2} + {y}^{4}}{6 {x}^{2} y - 4 x {y}^{3} - {x}^{4} - 1}$

#### Explanation:

Implicit differentiation is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. $y = f \left(x\right)$ - written explicitly as functions of $x$. However, some functions y are written implicitly as functions of $x$. So what we do is to treat $y$ as $y = y \left(x\right)$ and use chain rule. This means differentiating $y$ w.r.t. $y$, but as we have to derive w.r.t. $x$, as per chain rule, we multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Hence implicit differential of $- y = {x}^{4} y - 3 {x}^{2} {y}^{2} + x {y}^{4}$ is

$- \frac{\mathrm{dy}}{\mathrm{dx}} = \left(4 {x}^{3} y + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}}\right) - 3 \left(2 x {y}^{2} + {x}^{2} \times 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(1 \times {y}^{4} + x \times 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

or $- \frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} y + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} - 6 x {y}^{2} - 6 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{4} + 4 x {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}$

or $6 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} - 4 x {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} y - 6 x {y}^{2} + {y}^{4}$

or $\left(6 {x}^{2} y - 4 x {y}^{3} - {x}^{4} - 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} y - 6 x {y}^{2} + {y}^{4}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 {x}^{3} y - 6 x {y}^{2} + {y}^{4}}{6 {x}^{2} y - 4 x {y}^{3} - {x}^{4} - 1}$