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Though the following seems intuitive, is it correct precisely?

Suppose $T:\mathbb{R}^2\to\mathbb{R}$ is a smooth function. Then $$\Big({\partial\over\partial y}{\partial T\over\partial x}\Big)(0, y_0) = {d\over dy}\Big({\partial T\over\partial x}(0, y)\Big)(y_0)$$ for all $y_0\in\mathbb{R}$.

What I expect is to "see" the mathematical machinery that gets me from the LHS to the RHS.

Thanks!


Note: I'm a physics undergraduate, who is trying to understand the math (as precisely as possible) that the physics textbooks often sweep under the rug.

Also, to prevent abuse of notation, I think that RHS is better written as $\mathcal{T}'(y_0)$, where $\mathcal{T}:\mathbb{R\to R}$ is a function defined by $\mathcal{T}(y):={\partial T\over\partial x}(0, y)$

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The meaning of the LHS is as follows: we first define $f: \Bbb{R}^2 \to \Bbb{R}$ to be the function $f:= \dfrac{\partial T}{\partial x}$. In other words, for every $(\alpha, \beta) \in \Bbb{R}^2$, \begin{align} f(\alpha, \beta) := \dfrac{\partial T}{\partial x}(\alpha, \beta) \end{align} So, we now want to express $\dfrac{\partial f}{\partial y}(0,y_0)$ in a different way. Well, by definition, \begin{align} \dfrac{\partial f}{\partial y}(0,y_0) &:= \dfrac{d}{dy} \bigg|_{y=y_0}\bigg( y \mapsto f(0,y)\bigg) \tag{$*$}\\ &:= \dfrac{d}{dy} \bigg|_{y=y_0}\bigg(y \mapsto \dfrac{\partial T}{\partial x}(0, y) \bigg) \end{align}

So, really, this equality is true by definition of a partial derivative. Now, at this point, you might find my first equality a little suspicious and wonder if it is really true by definition. Well, ok depending on how you look at it, it's either a definition or a simple theorem with a one-line-proof (so really, I just consider it a definition).


You're probably more familiar with this definition:

Let $f: \Bbb{R}^2 \to \Bbb{R}$ be a function, and $(\alpha, \beta) \in \Bbb{R}^2$ be a point. If the limit \begin{align} \lim_{h \to 0} \dfrac{f(\alpha, \beta + h) - f(\alpha, \beta)}{h} \end{align} exists, we denote it as $\dfrac{\partial f}{\partial y}(\alpha, \beta)$, or $(\partial_2f)(\alpha, \beta)$, and call it the partial derivative of $f$ with respect to the second variable, evaluated at the point $(\alpha, \beta)$.

But if you look carefully, what is this definition saying? Well, the existence of that limit is precisely equivalent to the differentiability of the function $y \mapsto f(\alpha, y)$ at the point $\beta$. In other words, the partial derivative (if it exists) is exactly equal to any of the following: \begin{align} \dfrac{\partial f}{\partial y}(\alpha, \beta) &:= \bigg( y \mapsto f(\alpha, y)\bigg)'(\beta) \\ &\equiv \dfrac{d}{dy} \bigg|_{y = \beta} \bigg( y \mapsto f(\alpha, y)\bigg)\\ &= \dfrac{d}{ds} \bigg|_{s = 0} \bigg( s \mapsto f(\alpha, \beta + s)\bigg)\\ &= \dfrac{d}{ds} \bigg|_{s = 0} \bigg( s \mapsto f\left((\alpha, \beta) + s e_2 \right)\bigg) \end{align} where the $\equiv$ means "same thing expressed in different notation", and the third line is true by a simple application of the single variable chain rule (or just look at the limit definition, it's trivially true), and $e_2 = (0,1)$.


BTW, you're perhaps right to introduce a new function $\mathcal{T}$ to avoid any abuse of notation, but I didn't feel like coming up with new names for each new function I defined along the way, which is why I used the stopped arrow $\mapsto$ notation to indicate precisely what the function is, and I was also careful to distinguish between the bound and free variables, and a function vs the function's values at a point so hopefully it's clear exactly how to read the above notation.

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