# Implicit differentiation in a nutshell

What is implicit differentiation? When we are dealing with derivatives of functions in calculus, we often encounter functions such as y = f(x) where some variable y can be explicitly expressed as a function of an independent variable x, so the differentiation process is straight-forward, the derivative of y is denoted as dy/dx or y’(x) and simply requires operating on f(x) using our standard rules of differentiation.

But not all functions can be written in this explicit form y = f(x). In fact, many times there are functions which cannot be separated that easily, for example g(x, y) = c is a common representation of “implicit functions”, where g(x, y) is a function of both x and y, but we know y is still a function of x and x is an independent variable.

So how can we deal with such situations? Does the derivative even exist? The short answer is: yes, it does, but the way we evaluate it is slightly different. Consider the following example:

this here is an implicit function where y cannot be isolated entirely. In fact, this has the form g(y) = f(x), which means that in order to obtain the derivative y’(x) we need to do some clever manipulation. Firstly, we recall a simple fact from differentiation: if y = f(x), and we have a function g(y), then the derivative of g(y) with respect to x has to be equal to the derivative of y with respect to x times the derivative of g with respect to y. This is called the chain rule and is expressed as:

Because the derivative is a “linear operator”, meaning that the derivative of a sum of terms is equal to the sum of derivatives; differentiating both sides of Equation 1 yields:

and then on the left-hand side, we use the chain rule for each of those functions of y, and on the right-hand side we just use regular differentiation techniques, so then we obtain

The final step here is to notice how on the left-hand side, we have a common term dy/dx, so re-arranging everything in terms of dy/dx yields:

If you look at this closely, this is nothing more than the derivative y’(x) that we wanted! Somehow, we were able to differentiate Equation 1, and obtain an expression in terms of both x and y. Of course, this means that to obtain the derivative of this function at some point (x0, y0) we need to actually put in two values, rather than one which is normally the case with expliciy functions, but nonetheless: it is doable.