How does implicit differentiation work?

Aug 5, 2014

Implicit differentiation is a way of differentiating when you have a function in terms of both x and y. For example:

${x}^{2} + {y}^{2} = 16$

This is the formula for a circle with a centre at (0,0) and a radius of 4

So using normal differentiation rules ${x}^{2}$ and 16 are differentiable if we are differentiating with respect to x

$\frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(16\right)$

$2 x + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 0$

To find $\frac{d}{\mathrm{dx}} \left({y}^{2}\right)$ we use the chain rule:

$\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

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

$2 x + 2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Rearrange for $\frac{\mathrm{dy}}{\mathrm{dx}}$

dy/dx=(-2x)/(2y

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

So essentially to use implicit differentiation you treat y the same as an x and when you differentiate it you multiply be $\frac{\mathrm{dy}}{\mathrm{dx}}$

Theres another video on the subject here