What is implicit differentiation?

1 Answer
Jan 1, 2016

I tried this hoping it is understandable!

Explanation:

You may remember the difference between an implicit and explicit form of a function:
Normally your function can be read explicitly as
$y = \text{something of x}$
and derived immediately to find $\frac{\mathrm{dy}}{\mathrm{dx}} = \text{something else of x}$;

For example $y = {x}^{2} - 3 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 3$

Sometimes your function is more difficult to "extricate" and can be difficult to write as $y = \text{something of x}$.
You can only write it as $y \left(x\right) + a {x}^{n} + \ldots c = 0$
As you can see the $y$ is nested inside the other bits and it is in itself a function of $x$.

Consider for example a circle:
${x}^{2} + {y}^{2} = 4$
Here $y$ is "difficult" to extract (ok, not impossible) and if you try to derive as it is, you have to remember to derive also $y$:
$2 x + \textcolor{red}{2 y \frac{\mathrm{dy}}{\mathrm{dx}}} = 0$
Rearranging:
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x}{2 y} = - \frac{x}{y}$
As you can see this derivative contains again $y$; once you can figure out the form of $y$ you could substitute it in and get the usual derivative containing only $x$.

In the example of the circle you can write it as (extracting $y$):
$y = \pm \sqrt{4 - {x}^{2}}$
So that gives you:
(dy)/(dx)=-x/(color(red)(+-sqrt(4-x^2))

It isn't always possible to extract $y$ and you leave the derivative as it is, with $y$ in it.

To test this result try to derive the explicit $y = \pm \sqrt{4 - {x}^{2}}$ form and see if it checks with the one we found implicitly.