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

Jan 8, 2016

To differentiate implicitly, simply differentiate each term with respect to $x$, then rearrange for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Answer:$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} y + 1}{8 y - {x}^{3} - 1}$

#### Explanation:

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

I will take it piece by piece for the sake of clarity.

$\frac{d}{\mathrm{dx}} \left(4 {y}^{2}\right) = 8 y \frac{\mathrm{dy}}{\mathrm{dx}}$

This is from applying the chain rule.

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

This is from applying the product rule.

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} \left(x\right) = 1$

Putting that all together we get:

$8 y \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {x}^{2} y + \frac{\mathrm{dy}}{\mathrm{dx}} + 1$

Rearrange:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(8 y - {x}^{3} - 1\right) = 3 {x}^{2} y + 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} y + 1}{8 y - {x}^{3} - 1}$

Sometimes it is also possible to find $y$ in terms of $x$ in the original equation and sub back to get $\frac{\mathrm{dy}}{\mathrm{dx}}$ only in terms of $x$.