6

I'm a high schooler and today my teacher taught us the implicit differentiation, in which he gave us a very brief explanation of implicit function.

I didn't quite get it at that time so I decided to look it up a little and honestly, the concept of implicit functions just doesn't make sense to me, why are they even functions? The definitions of functions I was taught almost totally contradicts everything about implicit functions.

Also, how does it make sense to differentiate something which is not even in the explicit form, $y= f(x)$, seeing as how the definition of derivative is given using exactly that,

$f'(x) = \lim\limits_{h\to0} \dfrac{ f(x+h)- f(x)}{h}$

But the main thing that concerns me is, if the implicit functions are of the form $F(x,y)=0$ (which I assume is the equation of all points lying on the curve formed on the XY plane when the surface intersects it) then why do we assume $y$ is some composite function of $x$ and find $\frac{dy}{dx}$?

I'm really really confused right now and I have lots and lots of questions, someone pls help.

EDIT : $f$ is a function from $A$ to $B$, if it maps every element of $A$ to a unique element of $B$

in which case,

$x^2+y^2 = 4$ fails the vertical line test.

Gregory J. Puleo
  • 6,763
  • 4
  • 25
  • 31
William
  • 5,151
  • It would help if you explained why you think the definition of functions you have been taught contradicts the idea of implicitly-defined functions. (I assure that it doesn't.) – saulspatz Jan 23 '19 at 18:54
  • 1
    I've slightly changed your title, because an "even function" is a thing unto itself, so the old title might have caused some confusion. – Gregory J. Puleo Jan 23 '19 at 18:56
  • @saulspatz done – William Jan 23 '19 at 18:57
  • @William it looks like you edited the question shortly after my title edit, rolling back the title -- I'm not sure whether this was intentional or not, and I'm not going to fight you if you prefer your old title, but I'd suggest considering the title I proposed to avoid confusion. – Gregory J. Puleo Jan 23 '19 at 19:07
  • @GregoryJ.Puleo I didn't, I only edited my definition of a function. You're right, that's a slightly misleading title. – William Jan 23 '19 at 19:14
  • @GregoryJ.Puleo please change it back – William Jan 23 '19 at 19:19
  • @William I suspect the chain of events here was: I started editing title -> you started editing body -> I finished editing title -> you finished editing body, and your editing the body caused my title edit to get squashed. – Gregory J. Puleo Jan 23 '19 at 19:33

3 Answers3

4

The relation $x^2+y^2=4$ is not a function globally. That is, if look at all points $(x,y)$ that satisfy the equation, we get cases where the pairs have the same first element. However, the relation defines a function locally. If we pick a point $(x,y)$ that satisfies the relation, (other than $(\pm2, 0)$) then for nearby points, we have $y$ as a function of $x$. For example, if $(x, y)=(\sqrt{2}, -\sqrt{2})$ then we can take $y=-\sqrt{4-x^2}$ for $x$ close to $\sqrt{2}.$

In this case, we can find an explicit formula, but in many cases, we wouldn't be able to. This is where implicit differentiation comes in handy.

EDIT

In response to the OP's comment, you get to choose which variables to solve for in terms of the others, provided only that the hypotheses about derivatives are satisfied. If you look at the wiki page it talks about a function $f:\mathbf{R}^{n+m}\to\mathbf{R}^n$ This means that we have $m$ equations in $n+m$ unknowns. Then is says we can find a function $g:U\to\mathbf{R}^m$ where $U$ is a subset of $\mathbf{R}^n$ This mean that we can "solve" for $m$ of the variables in terms of $n$ of them, although we usually can't actually produce the solution. Rather, we can say that a solution exists. In your example of a single equation in six variables, you can solve for any one of them in terms of the other five, if the hypotheses of the theorem are satisfied.

Here is a detailed example with two equations in four variables. In the example, he talks about the solution for $u$ and $v$ in terms of $x$ and $y,$ but he might have made some other choice. In particular if that function turns out to be one-to-one, it would also be possible to solve for $x$ and $y$ in terms of $u$ and $v$. I'm afraid you won't be able to follow all the computations, since you have studied calculus of functions of several variables yet, but I think you'll be able to get the idea.

saulspatz
  • 53,824
  • It's somewhat beginning to make sense, although, relation defines a function locally, can you elaborate a little on this? Also, by that logic, would it be alright to conclude $\dfrac{dx}{dy}$ for a function $y= f(x)$ would exist, even if it is not invertible? – William Jan 23 '19 at 19:35
  • 1
    "Locally" just means near some point. I mean that for values of $x$ in some interval about $\sqrt{2}$ we can define a function $g(x),$ such that $g(\sqrt{2})=-\sqrt{2}$ and $x^2+g^2(x)=4.$ As to your question about ${dy\over dx}$ one can conclude that the derivative exists, under rather mild assumptions, but I don't know how to state them without talking about partial derivatives. – saulspatz Jan 23 '19 at 21:32
  • hey I've been reading about implicit function, and IF theorem, and partial derivatives. And I think your answer now makes a little sense. However I still have a question, in the implicit function, how do you decide the dependent or independent variable? Let's say you have an implicit function 6 variables, how would I know the independent and dependent variables in this case? (Because without knowing that, I wouldn't know differentiate what wrt what) – William Jan 26 '19 at 14:05
2

You might change your mind by considering that every function can be written in the implicit form

$$F(x,y)=0$$ where $$F(x,y):=y-f(x).$$

Obviously, an equation like $F(x,y)=0$ is often multi-valued (several $y$ for one $x$), but one can split the curve in several mono-value pieces. Implicit functions are so important that one should not be stopped by this technicality.

0

There's no reason that the locus of points satisfying $F(x,y) = 0$ has to be a function in the sense of its graph passing the vertical line test: e.g. if $F(x,y) = x^2 + y^2 - 1$ then that locus would be a unit circle. However, some functions that really are functions, whose graphs pass the vertical line test, would be prohibitive to explicitly write down in the form $y = f(x)$, but might be more easily described as the set of points satisfying $F(x,y) = 0$.

As a contrived example, if you'd never heard of the $\ln$ function, you would have trouble turning the equation $x = e^y$ into an explicit function of $y$ in terms of $x$, but you could still implicitly define $f(x)$ for $x > 0$, and find its derivative.

Gregory J. Puleo
  • 6,763
  • 4
  • 25
  • 31