# How do you find (dy)/(dx) given 4x^2+4xy=-5x^3y+4?

Jul 14, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{15 {x}^{2} y + 8 x + 4 y}{4 x + 5 {x}^{3}}$

#### Explanation:

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

However, we only differentiate explicit functions of $y$ wrt $x$. But if we apply the chain rule we can differentiate an implicit function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Example:

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

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

Now, we have:

$4 {x}^{2} + 4 x y = - 5 {x}^{3} y + 4$

Implicitly differentiating wrt $x$ (applying product rule):

$8 x + \left(4 x\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(4\right) \left(y\right) = \left(- 5 {x}^{3}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(- 15 {x}^{2}\right) \left(y\right)$

$\therefore 8 x + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 4 y = - 5 {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - 15 {x}^{2} y$

$\therefore 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 5 {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = - 15 {x}^{2} y - 8 x - 4 y$

$\therefore \left(4 x + 5 {x}^{3}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \left(15 {x}^{2} y + 8 x + 4 y\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{15 {x}^{2} y + 8 x + 4 y}{4 x + 5 {x}^{3}}$

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 $F \left(x , y\right) = 4 {x}^{2} + 4 x y + 5 {x}^{3} y - 4$; Then;

$\frac{\partial F}{\partial x} = 8 x + 4 y + 15 {x}^{2} y$

$\frac{\partial F}{\partial y} = 4 x + 5 {x}^{3}$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{8 x + 4 y + 15 {x}^{2} y}{4 x + 5 {x}^{3}}$, as before.