First of all I'd like to say that English is not my native language but I hope I've translated most of the concepts correctly.

So I'm going over my multivariable calculus textbook (also not in English, obviously) and most of the stuff I've understood. All the concepts are presented with functions of two variables.

They've defined partial derivatives with the following notations:

$$f^{'}_{x}(x_0, y_0) = \lim_{h\to0}\frac{f(x_0+h, y_0)-f(x_0, y_0)}{h}$$ $$f^{'}_{y}(x_0, y_0) = \lim_{h\to0}\frac{f(x_0, y_0+h)-f(x_0, y_0)}{h}$$

So far so good. Then for the differentiability in general, they defined it as follows:

The function $$f$$ is differentiable at some point $$\mathbf{x_0} = (x_0, y_0)$$ if there is a linear mapping $$L: \mathbb{R^2} \rightarrow \mathbb{R}$$ such that $$f(\mathbf{x_0} + \mathbf{h}) = f(\mathbf{x_0}) + L\mathbf{h} + o(\mathbf{h})$$ when $$\mathbf{h} \to \mathbf{0}$$.

I don't want to get into the whole deal but essentially they write everything in coordinates, using $$\mathbf{h} = (h, k)$$ and the linear mapping then becomes $$L(h, k) = ah + bk$$ where we later prove that $$a$$ and $$b$$ are partial derivatives: $$a = f^{'}_{x}(x_0, y_0), \quad b = f^{'}_{y}(x_0, y_0)$$

And from there we got $$L(h, k) = f^{'}_{x}(x_0, y_0)h + f^{'}_{y}(x_0, y_0)k$$

Then comes the following paragraph which confuses me completely:

The linear mapping $$L$$ for which we've seen is of form $$L(h, k) = f^{'}_{x}(x_0, y_0)h + f^{'}_{y}(x_0, y_0)k$$ is called the differential or the derivative of the function f at the point $$(x_0, y_0)$$ and we use the notation $$df(x_0, y_0)$$. Here's the confusing sentence:

If we mark linear mappings $$(x, y) \to x$$ and $$(x, y) \to y$$ as $$dx$$ and $$dy$$ respectively we get $$df(x_0, y_0) = f^{'}_{x}(x_0, y_0)dx + f^{'}_{y}(x_0, y_0)dy$$

Where did that come from? Why do we even introduce those two mappings? Where did the $$h$$ and $$k$$ go?

• The transition from your equation with $L(h,k)$ to the equation in your yellow box may be confusing for you because the former equation shows the value (which is a real number) of the linear map $L$ at the vector $(h,k)$ whereas the latter equation is showing identity of linear operators, not identity of real numbers. Aug 7 '19 at 7:31
• What do you mean by 'the identity of linear operators' ?
– Koy
Aug 7 '19 at 7:32
• "Where did the $h$ and $k$ go?" The function $df(x_0, y_0)$ on the left is equal to the function on the right. If you were to plug $(h,k)$ as input into the function on the left, the output $df(x_0,y_0)(h,k)$ would be the same as the output you'd get if you plugged $(h,k)$ into the function on the right. Aug 7 '19 at 7:33
• The point is that $L$ is a linear mapping that sends the vector $(h,k)$ to the real number $f’_x(x_0,y_0)h+f’_y(x_0,y_0)k$. If you let $dx$ represent the linear mapping that projects a vector onto its first coordinate and $dy$ represent the linear mapping that projects a vector onto its second coordinate, you’ll see that the mapping $f’_x(x_0,y_0)dx+f’_y(x_0,y_0)dy$ is also linear and also maps the vector $(h,k)$ to the real number above. Hence the two linear mappings must be the same. Aug 7 '19 at 7:35
• As an analogy, consider the difference between writing $f(x)=x^2$ and $f=g$ for two real-valued functions $f$ and $g$. The former equation is a relationship between real numbers; it tells you the value of the function $f$ at the input $x$. The latter equation is a relationship between functions, not real numbers; it tells you the values of $f$ and $g$ agree everywhere. Aug 7 '19 at 7:41

"Where did the $$h$$ and $$k$$ go?"
Typically, the linear map $$L: \Bbb{R}^2 \to \Bbb{R}$$ can be denoted by the symbol $$df_{(x_0,y_0)}$$. (I put $$(x_0,y_0)$$ in subscript position only to make things easier to read; otherwise there will be a lot of brackets involved). I hope you understood the equality \begin{align} L(h,k) = f_x'(x_0,y_0) h + f_y'(x_0,y_0) k \end{align} Now, with the notation $$L= df_{(x_0,y_0)}$$, we can write this above equality as \begin{align} df_{(x_0,y_0)}(h,k) = f_x'(x_0,y_0) \cdot h + f_y'(x_0,y_0) \cdot k \tag{*} \end{align} Note that in this equation, the LHS is a real number, and the RHS is a real number. The only reason we got real numbers is because we evaluated everything at the point $$(h,k)$$.
Now, what is being suggested is to define two new linear transformations $$dx:\Bbb{R}^2 \to \Bbb{R}$$ and $$dy: \Bbb{R}^2 \to \Bbb{R}$$ by the rules \begin{align} dx(h,k) = h \quad \text{and} \quad dy(h,k) = k \end{align} In words, you feed a two-component vector into $$dx$$, and the result you get is the first component. Similarly, $$dy$$ tells you the second component of each vector. Now, we can rewrite the equation $$(*)$$ as follows: \begin{align} df_{(x_0,y_0)}(h,k) = f_x'(x_0,y_0) \cdot dx(h,k) + f_y'(x_0,y_0) \cdot dy(h,k) \tag{**} \end{align} This is once again an equality of real numbers. Now, we can write this equality as an equality between linear transformations as follows: \begin{align} df_{(x_0,y_0)} = f_x'(x_0,y_0) \cdot dx + f_y'(x_0,y_0) \cdot dy \tag{\ddot{\smile}} \end{align} So, here, we have omitted the arguments $$(h,k)$$ of the functions. Now, on the LHS, we have a linear transformation, while on the RHS, we have a sum of linear transformations, so that we have a linear transformation on both sides of the equality.
What happened above is we defined new functions to avoid writing the arguments $$(h,k)$$. Consider the following simple $$1$$-dimensional example. Define three functions $$\alpha, \beta, \gamma: \Bbb{R} \to \Bbb{R}$$ by \begin{align} \alpha(t) = t \quad \text{and} \quad \beta(t) = t^2 \quad \text{and} \quad \gamma(t) = t + t^2 \end{align}
It should be obvious that \begin{align} \text{for every t \in \Bbb{R}, \gamma(t) = \alpha(t) + \beta(t)} \end{align} This is an equal sign between real numbers. But we can also write this more compactly as an equality of functions as just \begin{align} \gamma = \alpha + \beta \end{align}