What is the product of magnitudes $\frac{\partial }{\partial x}$ and $x$?

Question

I know that $\frac{\partial}{\partial x}\space (x)=1$, here I am not talking about it. Consider:
$$(\widehat{e}_x\frac{\partial}{\partial x}).(\widehat{e}_x x)=(\frac{\partial }{\partial x})\space(x)\cos0=(\frac{\partial }{\partial x})\space(x)$$

Before knowing the final answer lets have some basic knowledge. According to ordinary multiplication, $\color{blue}{(A)(B)=C \implies A=\frac{C}{B}}$ .

Final answer:

In solving dot product we need to multiply the magnitudes of the vectors, the same we are doing here, the "multiplication" of the magnitudes $\frac{\partial}{\partial x}$ and x, but in the name of multiplication, differentiation is done! Is what is done here (differentiation) really multiplication?

$$\color{red}{(\frac{\partial}{\partial x})\space(x)=\frac{\partial x}{\partial x}=1}$$

If it (differentiation) is multiplication, is the ordinary multiplication condition satisfied in the above answer, i.e is $\color{blue}{\frac{\partial }{\partial x}=\frac{1}{x}}$? As I understand it is not satisfied, what do you all say? It seems some new multiplication rule has been introduced here.

Answer 1

Does $\sin \frac \pi 2 = 1$ imply $\sin = \frac 2 \pi$ ?

Answer 2

Let $f:\mathbb{R}^2\to \mathbb{R}$. Then \begin{equation} \frac{\partial}{\partial x}(f)=\lim_{h\to 0} \frac{f(x+h,y)-f(x,y)}{h}. \end{equation} This is the derivation of $f$ in the direction $x$. If $f(x,y)=x$ then we get \begin{equation} \frac{\partial}{\partial x}(x)=\lim_{h\to 0} \frac{x+h-x}{h}=1 \end{equation}

Answer 3

The first mistake is that we imagine $\hat e_x\frac{\partial}{\partial x}$ as an ordinary vector. Well, it isn't.

We can't write $(\widehat{e}_x\frac{\partial}{\partial x}).(\widehat{e}_x x)$ as the product of the magnitudes of $(\widehat{e}_x\frac{\partial}{\partial x})$ and $(\widehat{e}_x x)$ and the cosine of their angle. But we must firstly calculate how partial derivative acts on $(\widehat{e}_x x)$. After we found it, we have two vectors and dot product is applicable.

Answer 4

$(\frac{\partial }{\partial x})(x)$ is NOT multiplication of $\frac{\partial }{\partial x}$ by $x$. It is the application of the partial derivative operator $\frac{\partial }{\partial x}$ to the function $x$. The partial derivative operator takes in one function, in your case $x$, and produces another function, in your case the constant function $1$.

Talking about multiplication of partial derivative operator by a function makes as much sense here as talking about changing the tires on a horse.

Perhaps it is the notation that is misleading you. $\frac{\partial }{\partial x}$ is not $\partial\div(\partial\times x)$ any more than the $\sin$ function is the product of $s,i,n,$. It's just notation for the partial derivative operator with respect to $x$. In some cases when we are working with partial derivatives, it looks like we are working with fractions, but we are not. It only looks that way.