When we have something like $y = 2x$ we understand $y$ to be the value of the function $f$ at each point $x$ where $f(x) = 2x$, to reiterate, $y$ is not a function but merely a label for the output of $f$. In the case of implicit differentiation we may have something in the form $f(x,y) = 0$ where $f$ is the function and $x, y$ are variables over some domain. I have seen in a lot of introductory calculus books that they write, "to differentiate $f(x,y) = 0$ we use the chain rule and treat $y$ as a function of $x$", now I know that $\frac{dy}{dx}$ is not $0$ in these cases as when we change $x$ there is a resulting change in $y$, i.e the input variables of the function influence each other, but I would like to ask how is it valid to call $y$ as a function of $x$ when it clearly isn't? In addition to this, how does the implicit function theorem come into play here?
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1Just write f(x,y(x)) instead and it will be clear. – trula Aug 03 '23 at 13:07
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1you're caught over semantics, y and f(x) are used interchangeably in this context. in the implicit case, y is said to be a function of x because a slight change in x implies a slight change in y. – Colver Aug 03 '23 at 13:09
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1The first point is that it sometimes is valid: if $f(x,y) = 2x-3y+1$, then every value of $x$ determines a unique value of $y$, which is basically the definition of "function". – JonathanZ Aug 03 '23 at 13:09
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I think its most clear to write $f(x,g(x))$ where $y=g(x)$ then its clear $y$ is the label given to the function $g$ evaluated at $x$. Rigour is best. – Nav Bhatthal Aug 03 '23 at 13:10
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The comment about $g(x)$ is an interesting point. Would you consider adding it to the question? – David K Aug 03 '23 at 13:23
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Upvote interesting stuff. – Nav Bhatthal Aug 03 '23 at 13:28
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1Look up the Implicit Function Theorem. The proof of that contains the rigor that makes implicit differentiation work. – John Douma Aug 03 '23 at 14:13
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I have looked it up, and have one question, the function $g$ can be different for each interval right? – Nav Bhatthal Aug 03 '23 at 16:07
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1Not "different on each interval", but "different near each point". Consider $f(x,y) = x^2 + y^2 -1$. On the interval $x \in (-1/2, 1/2)$ there are different functions near $(x,y) = (0,1)$ and $(x,y) = (0,-1)$. – JonathanZ Aug 03 '23 at 16:17
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How do we conclude that we are free to differentiate $y$ at all points then? Because each unique $g$ is fully differentiable? Does that make $y$ a sort of piecewise function? (if such function exists) – Nav Bhatthal Aug 03 '23 at 16:30
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If we do not consider "$y$ a function of $x$" then it makes no sense at all to consider $dy/dx$ since the derivative is only defined for functions – FShrike Aug 03 '23 at 20:00
1 Answers
I had a similar question recently and here's how I understand it.
The implicit function theorem for two dimensions states that for a function
$$F:\mathbb{R}^2\rightarrow \mathbb{R}$$
$$(x,y)\mapsto F(x,y)$$
such that $F$ is continuously differentiable near $(x_0,y_0)$. If $F(x_0,y_0)=0$ and $(\partial_2F)(x_0,y_0)\neq0$, then there exists an open neighbourhood $U$ of $x_0$ and a function $$f:U\rightarrow\mathbb{R}$$That is continuously differentiable such that $f(x_0)=y_0$, $F(x,f(x))=0$ for all $x\in U$ and $f'(x_0)=-\frac{(\partial_1F)(x_0,y_0)}{(\partial_2F)(x_0,y_0)}$.
(Note that $\partial_iF$ stands for the partial derivative of F with respect to it's i-th argument.)
This is why we can treat $y$ as a function of $x$ even though it might be impossible to get an equation of the form $y=g(x)$, the graph of the curve clearly fails the vertical line test, etc. It's just that near every point on the curve, there exists some function whose graph will be the same as the graph of $F(x,y)=0$ near that point and $y=f(x)$ so we can use chain rule to differentiate. (Unless $(\partial_2F)(x_0,y_0)=0$ of course, in which case, there will be a vertical tangent so the derivative is undefined.)
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1The statement you made wasn't quite right, I've amended it. You can actually strengthen it, saying that further we can choose $U$ so that there is an open neighbourhood $V$ of $y_0$ such that $f(U)\subseteq V$ and $F(x,y)=0$ for $x\in U,y\in V$ iff. $y=f(x)$. That gives you the characterisation of the level set $F^{-1}{0}$ being locally equal to the graph of $f$ – FShrike Aug 03 '23 at 19:54
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You're absolutely right. As for the strengthening, I hope that anyone seeing this looks at your comment. – CroW Aug 04 '23 at 04:38
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Thanks for the comment, and as we know that there exists some $g$ near to each point on $F$ that implies that $\frac{dy}{dx}$ exist at all point on $F$ such that there is no vertical tangent), which allows us to use the chain rule when differentiating, even though $g$ is only local to each point? i.e there could be infinitely many $g's$ for each $F$? – Nav Bhatthal Aug 04 '23 at 09:16
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@NavBhatthal Yes, a simple example is $y^2=x$. Depending on the point $(x,y)$ we're evaluating the derivative at, the $f$ in $y=f(x)$ supplied by the theorem is different ($f(x)=\sqrt{x}$ for $y>0$ and $f(x)=-\sqrt{x}$ for $y<0$). I'm not really sure about the infinite part but yes there could be a large number of different $f$s for different points on the curve. (Also, while in this example we can find explicit formulae for $f(x)$ for different $(x,y)$, generally this doesn't need to be the case.) – CroW Aug 04 '23 at 09:38
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I guess I was getting confused with the (incorrect) idea that we can only find $\frac{dy}{dx}$ if there is one $f$, but the idea that multiple $f's$ exist is backed by the theorem, so all is good. – Nav Bhatthal Aug 04 '23 at 10:10
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@NavBhatthal This may or may not clarify things, but a point to remember is that while the particular function $f$ represents may be different for different $(x_0,y_0)$, when you differentiate say $x_0^2+y_0^2-1=0$ using $y_0=f(x_0)$ from IFT, you get $f'(x_0)=-\frac{x_0}{f(x_0)}$ so regardless of what $f$ actually is near any point, the value of $f$ at $x_0$ is what will be in the denominator, which from the theorem, will be $y_0$. So the result you get will work for all (valid) points on the curve. – CroW Aug 04 '23 at 10:48
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