# How do you implicitly differentiate xy^2-6x-8y=3?

Mar 6, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 - {y}^{2}}{2 x y - 8}$

#### 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:

$x {y}^{2} - 6 x - 8 y = 3$

Differentiate wrt $x$ (applying product rule):

$\left(x\right) \left(\frac{d}{\mathrm{dx}} {y}^{2}\right) + \left(\frac{d}{\mathrm{dx}} x\right) \left({y}^{2}\right) - \frac{d}{\mathrm{dx}} 6 x - \frac{d}{\mathrm{dx}} 8 y = \frac{d}{\mathrm{dx}} 3$
$\therefore \left(x\right) \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(1\right) \left({y}^{2}\right) - 6 - 8 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2} - 6 - 8 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore \left(2 x y - 8\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 6 - {y}^{2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 - {y}^{2}}{2 x y - 8}$

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 $F \left(x , y\right) = x {y}^{2} - 6 x - 8 y = 3$; Then;

$\frac{\partial F}{\partial x} = {y}^{2} - 6$

$\frac{\partial F}{\partial y} = 2 x y - 8$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{y}^{2} - 6}{2 x y - 8} = \frac{6 - {y}^{2}}{2 x y - 8}$, as before.