So I've just started to get into some calculus and I recently came across the topic of implicit differentiation. I am extremely confused on what implicit functions are and there is very little information on what exactly implicitly defined functions really are but instead lots of defintions on how to implicitly differentiate.

I became even more confused from this definition proofwiki provides which defines an implicit function as follows.

Consider a (real) function of two independent variables $z=f(x,y).$

Let a relation between $x$ and $y$ be expressed in the form $f(x,y)=0$ defined on some interval $I$.

If there exists a function:

$y=g(x)$ defined on $I$ such that:

$∀x∈I:f(x,g(x))=0$ then the relation $f(x,y)=0$ defines $y$ as an implicit function of $x$.

I can't seem to wrap this definition i am familiar with how a function is a special kind of binary relation but when we state that let a relation between $ x, y$ be expressed in the form $f(x,y)=0$ what exactly do we mean by this for example how is $x^2 +y^2=1$ a relation in that form. a simple example would be greatly appreciated.

I hope this question is not too vague to be answered or too silly to be even considered thanks in advance.

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    $\begingroup$ $x^2+y^2=1$ if and only if $x^2+y^2-1=0$ $\endgroup$ Apr 25, 2021 at 16:14
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    $\begingroup$ When you say "$f(x,y)=0$ defined on some interval $I$" I don't know what you mean. $\endgroup$ Apr 25, 2021 at 16:16
  • $\begingroup$ @ancientmathematician that is the definition given here google.com/url?sa=t&source=web&rct=j&url=https://… $\endgroup$ Apr 25, 2021 at 16:25
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    $\begingroup$ Thanks, I don't think that is very satisfactory - as you have found! $\endgroup$ Apr 25, 2021 at 16:38
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    $\begingroup$ @ancientmathematician Yes you're right, it's an appalling howler. This came from ProofWiki. Problem was, I was trying to make sense of some appallingly loose definitions in the sources I had. Yes of course it doesn't make sense to define an interval as a subset of the plane. I need to be kicked all around the playground for that. I've just amended that ProofWiki definition, but I'd be interested to read a proper rigorous definition rather than "it's something a bit like this" which is what you usually get. $\endgroup$ Apr 27, 2021 at 16:25

3 Answers 3


The main idea is that you have an equation that in theory could be solved for $y$ as a function of $x$ but the actual procedure is too hard/unknown. In other words, the equation gives enough information to uniquely specify a function but doesn't actually give it.

For example, $x^3y^3 + xy = 4$ is an equation cubic in $y$. With a lot of tedious work in algebra, it is possible to solve it and get $y$ alone. But if the equation is fifth degree or higher, then depending on the coefficients, there might not be any known way to find the formula directly.

It's a kind of specifying an answer by giving a problem that it solves. We call it indirect because it isn't "$y = $(blah that includes $x$'s but doesn't include any $y$'s)".

  • $\begingroup$ my confusion arises with why can we say that equations which clearly cannot be written explicitly in terms of $x$ then be considered equations of $y$ in terms of $x$. Like for quantic equations with $x,y$ it is impossible to ever express $y$ explicitly in terms of $x$ yet we blindly assume we can and differentiate anyway why is this? $\endgroup$ Apr 25, 2021 at 16:55
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    $\begingroup$ @Thehomeschooler It's still a function as long as we know that there's exactly one $y$ value for each $x$ value (which sometimes means we need to look at a subset of the possible $x$ and/or $y$ values). Writing an expression in terms of a variable with addition, multiplication, $sin$, $log$, etc. is one useful way of forming a function, but that's not what "function" really means. $\endgroup$
    – aschepler
    Apr 25, 2021 at 18:23
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    $\begingroup$ One way to prove the function exists is to pick a value of $x$, which gives you an equation to solve for $y$, and simply plugging in different values for $y$ to check if it is too big or small, and home in on the answer. So long as there is some number you are approaching, the function is defined at that point. Do that at a bunch of points, and you end up with a rough sketch. Numerical approximation is rough but with computers these days it is a common method and to my mind very concrete and direct. So long as you can home in on a $y$ for each $x$ then you have a function. $\endgroup$ Apr 25, 2021 at 21:27
  • $\begingroup$ @RobertTheTutor I am still a bit confused on how for example equations that are impossible to express y explicitly in terms of x can still be differentiated how does the implicit function theorem help us here as if there is no function $y=$(some x terms) then how can we hope to differentiate $\endgroup$ Apr 26, 2021 at 19:19
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    $\begingroup$ It's more like starting from an equation that $y$ has to satisfy, but can't necessarily be solved, and turning it into an equation that $y'$ has to satisfy, but can't necessarily be solved. Remember, even after the implicit differentiation, solving for $y'$ often will still give you a mess containing both $x$'s and $y$'s. There are parts of the equation we don't know, but other parts we do know, so we leverage our knowledge of those pieces as best we can. $\endgroup$ Apr 26, 2021 at 19:44

Here is what I think the story is.

Step (i) Suppose we have two sets $X,Y\subseteq\mathbb{R}$ and a function $f:X\times Y\to\mathbb{R}$.

Step (ii) This gives us a relation on $X\times Y$: define $x\sim y$ to mean $f(x,y)=0$.

[Note: In fact there's no need to introduce the idea of this relation.]

Step (iii) Suppose that we are in this situation, and that on some interval $I\subseteq X$ we have a function $g:I\to\mathbb{R}$ such that for all $x\in I$ it is true that $f(x,g(x))=0$. Then we say that $f$ defines $g$ implicitly on $I$.

[Note: We sometimes then talk about $g$ being an implicit function - this is loose, informal, talk. More carefully, $g$ is a function, which is implicitly defined on $I$ by $f$.]


There are already excellent answers, e.g. RobertTheTutor's that, in summary, say: the equation $f(x,y)=0$ defines a function $y=g(x)$ by solving for $y$ in theory, although in practice it is difficult / impossible / inconvenient to solve.

As per the comment my confusion arises with why can we say that equations which clearly cannot be written explicitly in terms of $𝑥$ then be considered equations of $𝑦$ in terms of 𝑥 and also the OP there is very little information on what exactly implicitly defined functions really are but instead lots of definitions on how to implicitly differentiate, here is my interpretation.

The definition of implicit function does not mention the derivative. But it turns out that the most useful way to prove that such implicit function exists, is the implicit function theorem which does use Calculus. There are excellent rigourous proofs in Calculus books but, informally, think that if that function $f$ exists and it is differentiable, then it should be possible to differentiate implicitly the equation $f(x,y)=0$ and solve for $y'$, that necessarily must give $y'=g'(x)$.

So here you are your answer: even if the equation cannot be written in practice explicitly in terms of $𝑥$, it can be considered in theory that it defines a function of $y$ in terms of $x$, if the conditions of the IFT hold, which can be checked by computing the implicit derivative.

EDIT: This question reminds us the very important fact that the IFT is local, i.e. it only asserts the existence of the function implicitly defined in a neighbourhood of some point where you have computed the implicit derivative.

EDIT 2: I think an important source of confussion here is the difference between what absolutely cannot be done and what I cannot do because it is not simple or immediate for me. Imagine I have not yet studied square roots. Then $y^2=x$ cannot be solved in terms of simple (for me) functions. But if a draw this curve near $(x,y)=(1,1)$, it turns out that it "looks like" a function (the vertical line test holds) so I can define ex novo the function $f$ such that its graph matches the curve $y^2=x$. And then, voilà, I am magically able to solve for $y=f(x)$! In other words, $f$ is defined as: given $x$, draw a vertical line that intersects the curve at a single point, which is by definition $y=f(x)$.

  • $\begingroup$ I am terribly confused i understand that some implicit equations i.e quintic equations are impossible to express y explicitly in terms of x so how does this theorem allow us to conclude there is?. $\endgroup$ Apr 26, 2021 at 19:16
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    $\begingroup$ @Thehomeschooler Quintic equations cannot be expressed in terms of simple functions. But if you can draw the curve and it "looks like" a function, then it can be expressed as a function, just it is a strange function that does not have still a name. $\endgroup$
    – Miguel
    Apr 27, 2021 at 13:58
  • $\begingroup$ @Thehomeschooler - Think about sin(x) and cos(x) - there is NO WAY to write these in terms of polynomials, but they are still functions. We just gave them names! Likewise, for some weird function, I can just invent the function name q(x) to represent the solution! There may not be any polynomial that represents that. The question is simply whether or not there is a logical relationship that exists. $\endgroup$
    – johnnyb
    Apr 27, 2021 at 14:30
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    $\begingroup$ An additional problem presents itself in that "functions" are often defined as having only one value for any given input. Many "implicit functions" are not like that. For instance, a circle is usually considered an implicit function, even though for every point there are two $y$ values for each $x$. However, as @Miguel points out, the implicit function theorem allows us to treat something as an implicit function, even if it is only a function within a certain neighborhood. So, if we zoom in enough to $x^2 + y^2 = C$, it will look like a function where we are looking. $\endgroup$
    – johnnyb
    Apr 27, 2021 at 14:32
  • $\begingroup$ By the way, if you are homeschooling, I actually wrote a calculus book with homeschoolers in mind - "Calculus from the Ground Up". It is thorough (actually allows you to solve more problems than most intro calc books), but not as technical as you seem to be looking at (I try to justify everything intuitively even if not explicitly). $\endgroup$
    – johnnyb
    Apr 27, 2021 at 14:35

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