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

Feb 24, 2017

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

#### 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:

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

Differentiate wrt $x$:

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

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

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

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

$\frac{\partial F}{\partial x} = - 2 x \setminus \setminus \setminus$ and $\setminus \setminus \setminus \frac{\partial F}{\partial y} = 1 + 3 {y}^{2}$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{- 2 x}{1 + 3 {y}^{2}} = \frac{2 x}{1 + 3 {y}^{2}}$, as before.