# Pulling a partial derivative out

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 $$\mathcal{T}(y):={\partial T\over\partial x}(0, y)$$

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.