Suppose you do a transformation of variables from (x, a)-> (y, b) like from cartesian to polar.I know that $$\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}$$ This holds true for total derivatives but not for partial derivative. Does anybody has any logical explanation/intuition behind this?
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Where did you get that it does not hold true for partial derivatives, which are just the usual single derivatives?
For instance, if you now let $y=f(z,w),$ and moreover if as a function of $z$ alone we have that $y$ is continuously differentiable and $y_z\ne 0,$ then the derivative of the inverse $y\mapsto z$ exists, and is still given by $$\frac{1}{\frac{\partial y}{\partial z}}.$$
Allawonder
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See https://math.stackexchange.com/questions/1744014/partial-derivatives-inverse-question for a more elaborate explanation regarding why it doesn't hold in partial derivatives. – Rishabh Jain Oct 05 '19 at 19:45
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1@RishabhJain I don't see any where in there where a function satisfying my conditions above fails to fulfil the consequences claimed. – Allawonder Oct 05 '19 at 19:52
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In the post I have mentioned, Mark Viola gave an example where reciprocal of a partial derivative is not it's inverse. – Rishabh Jain Oct 05 '19 at 19:56
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@RishabhJain If $x=r\cos\phi,$ where $\phi$ is fixed, then for all points where we have that $\partial x/\partial r=\cos\phi.$ In any domain of $r$ where this has the same (nonzero) sign, the inverse $r\mapsto x$ exists and its derivative is given by $\sec\phi.$ – Allawonder Oct 05 '19 at 20:03
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@RishabhJain If $x=r\cos\phi,$ where $\phi$ is fixed, then for all points $r$ where we have that $\partial x/\partial r=\cos\phi$ has the same (nonzero) sign (i.e., any fixed $\phi\ne (2k+1)π/2$), the inverse $r\mapsto x$ exists and its derivative is given by $\sec\phi.$ – Allawonder Oct 05 '19 at 20:04
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@Allawonder: The thing is that the notation $\partial r/\partial x$ is usually implicitly taken to mean the partial derivative of $r(x,y)=\sqrt{x^2+y^2}$ holding $y$ (the “partner variable” of $x$) fixed, not holding $\phi$ fixed. And that is definitely not equal to $1/(\partial x/\partial r)$ (where the derivative is taken holding $\phi$ fixed). So in that sense the statement is false for partial derivatives. – Hans Lundmark Oct 05 '19 at 20:54
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@HansLundmark Even if that's how $r$ is defined, we have that $r_x=x/r,$ and in any neighborhood where this derivative preserves its sign and doesn't vanish (e.g., for $x>0$), it is still the case that the derivative of the inverse $x\mapsto r$ exists and is given by $r/x.$ – Allawonder Oct 06 '19 at 05:19
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Yes, of course, if you're keeping the same variable fixed for both the forward and the backward map. But then it's just single-variable calculus, really. I'm talking about the map $(r,\phi) \mapsto (x,y)$ and its inverse $(x,y) \mapsto (r,\phi)$. Then it's the whole Jacobian matrices that are mutual inverses, not individual partial derivatives. For example, $\partial r/\partial x$ with $y$ fixed equals $x/r=\cos \phi$ (since $r(x,y)=\sqrt{x^2+y^2}$), and $\partial x/\partial r$ with $\phi$ fixed also equals $\cos \phi$ (since $x(r,\phi)=r \cos\phi$)! In that sense the statement is false. – Hans Lundmark Oct 06 '19 at 10:52
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@HansLundmark But that's exactly the point I've been making since (see my answer above again) -- we are indeed keeping the same variable fixed; why would we suddenly change it? And yes, as I explained in my answer above, that's just single-variable calculus. So the map I was talking about was (see my answer again) $y\mapsto z,$ where $y=f(z,w),$ for fixed $w,$ for example. – Allawonder Oct 06 '19 at 13:06
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Why change which variable that's fixed? Well, because that's what people almost always mean when they use this notation! The notation $\partial f/\partial x$ is ambiguous, since it doesn't say what should be kept fixed as $x$ varies, but the usual interpretation is that $x$ is one of the variables in a coordinate system $(x,y)$, and it's the partner variable $y$ that should be kept fixed. But when one writes $\partial f/\partial r$, it's understood that $f$ is expressed in polar coordinates $(r,\phi)$, and then the context dictates that it is $\phi$ that is to be held fixed. (Not $y$!) – Hans Lundmark Oct 06 '19 at 13:49
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And since the question mentions that "it's not true for partial derivatives", I would surely think that whoever said that to the OP was referring to situations such that the one I've described (it's a standard example in multivariable calculus courses). – Hans Lundmark Oct 06 '19 at 13:51
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@HansLundmark I don't understand what you're talking about. My question was, suppose we have a function $f(x,y)=z.$ Then holding, say $y$ fixed, this question is about, and I answered about, the map $x\mapsto z$ and its possible inverse. Why should this inverse be anything other than $z\mapsto x,$ if it exists? Well, if it's anything other than that, then you're talking of something else in any case. This something else is not what the question is about, nor what my answer is about. – Allawonder Oct 06 '19 at 14:27
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And I don't understand what you're talking about... The question specifically says "Suppose you do a transformation of variables [...] like from cartesian to polar". That's a map from $\mathbf{R}^2$ to $\mathbf{R}^2$, not something where you just keep one variable fixed and do a coordinate transformation from $\mathbf{R}$ to $\mathbf{R}$ with the other variable. – Hans Lundmark Oct 06 '19 at 14:46
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@HansLundmark The question has been edited since. It seems to me that the question had been specifically about partial derivatives. – Allawonder Oct 06 '19 at 14:53
$$\frac{1}{\frac{dy(x)}{dx}} = \left.\frac{dx(y)}{dy}\right|_{y = y(x)}$$
– the_candyman Oct 05 '19 at 18:45