There is a passage in the Chapter 12 exercises of Spivak's Calculus (which is a book specific to real-valued functions) that reads as follows:
In general, determining on what intervals a differentiable function is defined implicitly by a particular equation may be a delicate affair, and is best discussed in the context of advanced calculus. If we assume that $f$ is such a differentiable solution, however, then a formula for $f'(x)$ can be derived by differentiating both sides of the equation defining $f$ (a process known as "implicit differentiation").
I would like to make sure I am correctly understanding this comment.
Consider two different expressions:
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$[-f(x)]^5-f(x)=x$
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$[f(x)]^3-3f(x)=x$
To spare some space, rather than derive statements about these expressions, consider their graphs of solutions (respectively) when we rewrite the expressions as: $-y^5-y=x$ and $y^3-3y=x$.
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In the first expression, we have that $f$ is a function on all of $\mathbb R$. In the second case, the graph reveals that $f$ is not a function on all of $\mathbb R$.
If I proceed with the implicit differentiation of 1., then I have:
$$f'(x)=\frac{-1}{1+5[f(x)]^4}$$
If I proceed with the implicit differentiation of 2., then I have:
$$f'(x)=\frac{1}{3[f(x)]^2-3}$$
In the first case, there is clearly a single $f$ defined on all of $\mathbb R$ that exists. So differentiating the expression implicitly strikes me as an acceptable 'cheat code' of sorts…to get around the difficulty of isolating $f$ to one side of the inequality.
In the second case, however, it is not apparent to me which function $f$ I am implicitly differentiating. Is this what Spivak means in his passage when he writes, "If we assume that $f$ is such a differentiable solution…"? Under these circumstances, what exactly does it mean, logically, to assume that such an $f$ is a solution to the expression and can therefore be differentiated?
It is simply: "Suppose $f_1$ is one such solution to the expression (where $f_1$ may necessarily be defined on a subinterval of $\mathbb R$) and is also differentiable. Then its derivative can be found through the implicit differentiation approach"
Best Answer
This is what we call a "necessary condition". If $S \implies N$, then $N$ is a necessary condition for $S$, because it's impossible for $S$ to be true without $N$ also being true, but if $S$ is false then the truth of $N$ is unknown.
In this case, $S = $ "$f$ is a differentiable function satisfying the given equation" and $N = $ "The derivative of $f$ satisfies the equation that results from implicit differentiation". So if a function $f$ does indeed exist, then it is guaranteed that its derivative meets that criterion.
Indeed, as you show one of the equations does not define a unique function on $\mathbb{R}$, but you can define multiple functions on different domains to represent various parts of the relation, and each of those functions will still have matching derivatives.
For reference, the opposite of a necessary condition is a "sufficient condition". If $S \implies N$, then $S$ is a sufficient condition for $N$ because knowing that $S$ is true is enough to know that $N$ is true, but it's also possible for $N$ to be true without $S$. For example, "$f$ is a differentiable function" is a sufficient condition for "$f$ is a continuous function". If a condition is both sufficient and necessary, then we might say that it is equivalent, i.e. $A \implies B$ and $B \implies A$ means $A \iff B$.