# How do you find (dy)/(dx) given -2xy^2-3x^2y^3+3=4x^3?

##### 1 Answer
Feb 24, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{12 {x}^{2} + 2 {y}^{2} + 6 x {y}^{3}}{4 x y + 9 {x}^{2} {y}^{2}}$

#### Explanation:

When we differentiate $y$ wrt $x$ we get $\frac{\mathrm{dy}}{\mathrm{dx}}$.

However, we cannot differentiate a non implicit function of $y$ wrt $x$. But if we apply the chain rule we can differentiate a function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

When this is done in situ it is known as implicit differentiation.

We have:

$- 2 x {y}^{2} - 3 {x}^{2} {y}^{3} + 3 = 4 {x}^{3}$

Differentiate wrt $x$ (applying product rule):

$\left(- 2 x\right) \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(- 2\right) \left({y}^{2}\right) + \left(- 3 {x}^{2}\right) \left(3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(- 6 x\right) \left({y}^{3}\right) = 12 {x}^{2}$

$\therefore - 4 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {y}^{2} - 9 {x}^{2} {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 6 x {y}^{3} = 12 {x}^{2}$

$\therefore - \left(4 x y + 9 {x}^{2} {y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 12 {x}^{2} + 6 x {y}^{3} + 2 {y}^{2}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{12 {x}^{2} + 6 x {y}^{3} + 2 {y}^{2}}{4 x y + 9 {x}^{2} {y}^{2}}$

Advanced Calculus

There is another (often faster) approach using partial derivatives. Suppose we cannot find $y$ explicitly as a function of $x$, only implicitly through the equation $F \left(x , y\right) = 0$ which defines $y$ as a function of $x , y = y \left(x\right)$. Therefore we can write $F \left(x , y\right) = 0$ as $F \left(x , y \left(x\right)\right) = 0$. Differentiating both sides of this, using the partial chain rule gives us

 (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))

So Let $F \left(x , y\right) = 4 {x}^{3} + 2 x {y}^{2} + 3 {x}^{2} {y}^{3} - 3$; Then;

$\frac{\partial F}{\partial x} = 12 {x}^{2} + 2 {y}^{2} + 6 x {y}^{3}$

$\frac{\partial F}{\partial y} = 4 x y + 9 {x}^{2} {y}^{2}$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{12 {x}^{2} + 2 {y}^{2} + 6 x {y}^{3}}{4 x y + 9 {x}^{2} {y}^{2}}$, as before.