Implicit derivatives

A more suitable title for this section is “what can we say about the derivative of an a function defined implicitly?”

With that, the first step is to understand what is an implicit function, only then we bother with its derivative.

Implicit functions

An implicit function is a function defined by an equation. An example shows well the idea:

Consider the equation \(x^2 + y^2=1\). The truthset of this statement is the circle of radius one:

\[ \{(x,y)\in\mathbb{R}^2\,\,|\,\,x^2+y^2=1\} \]

We can use it to define functions, after all functions are also sets of ordered pair albeit with the property that each \(x\) can only appear one time in the set (please note this is not the case for the circle, each \(x\) occurs twice.)

What we are going to do is to slice, restrict, the set in such a way as a function comes out. There are many ways of doing so, but here is the more useful ones: Make the upper half of the circle one function and the bottom half another.

In mathematical notation, the upper half is defined as:

\[ \begin{align} f: &[-1,1] \longrightarrow \mathbb{R}\\ &x\longmapsto f(x):=[\text{positive $y$ solution of $x^2+y^2=1$}] \end{align} \]

the bottom half is:

\[ \begin{align} g: &[-1,1] \longrightarrow \mathbb{R}\\ &x\longmapsto g(x):=[\text{negative $y$ solution of $x^2+y^2=1$}] \end{align} \]

The functions \(f\) and \(g\) are defined implicitly, because the procedure that one must follow to assign \(y\) to the chosen \(x\) is implict in solving an equation, in this case \(x^2 + y^2=1\); by implicit I mean, the sequence of steps you must follow to solve it, is not given to you explicitly.

A more practical way to look at the upper half (similarly for the bottom half) is to say that \(y\) is the positive function of \(x\) such that it satisfies the equation \(x^2 +y^2=1\), and to make that evident we can write:

\[ x^2 +y(x)^2=1 \]This notation tell us: \(y\) is a function of \(x\), though how we get the value of \(y(x)\) from \(x\) requires to solve the equation with whatever means are necessary. It is precise this “with whatever means…” that promotes the \(y(x)\) function (equivalently the \(f\) function) to the class of implicit functions.

Derivative of an implicit function

We can almost antecipate, since the function is not given explicitly, chances are, its derivative is also not given explicitly.

Please note that when a function is given explicitly, we used the derivatives rules to obtain an explicit derivative function.

When dealing with implicit function, i.e., functions on which we only know the statement - in the case above \(x^2+y^2=1\) - that it obeys, the same thing occurs with its derivative, we’ll only know a statement it obeys.

The way to obtain that statement is to use the chain rule.

Take the derivative of the lhs and rhs of \(x^2 +y^2=1\), the result is:

\[ (x^2+y^2)'=1' \implies 2x+2y(x)y'(x) = 0 \]

On the rhs we find the statement: whatever the function \(y'(x)\) is, it obeys that equation, hence to find what is the value of \(y'(x)\) corresponding to the chosen \(x\), we have to solve the equation \(2x+2y(x)y'(x) = 0\). To do that, chose an \(x\), compute \(y(x)\) using whatever means necessary (the function \(y(x)\) is implicit!), plug in that result in the equation above and in turn solve it for \(y'(x)\).

As you can see the sequence of steps that one must follow to arrive at one evaluation is implicit!