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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.

Sensebe
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    What you have here is overloaded symbols. The () symbol is used for both multiplication and for the input of an operator. The partial differential operator is just that, an operator....so you don't have multiplication going on, and thus no such division – Alan Nov 21 '14 at 13:39
  • @Alan: You are supposed to do multiplication not to input anything to the operator. – Sensebe Nov 21 '14 at 13:44
  • @Ant: Yes, I agree with you. How does that answer my question? – Sensebe Nov 21 '14 at 13:52
  • -1 for skipping the comments that help you and only replying to the ones that confuse you. – djechlin Nov 21 '14 at 13:56
  • @Ant: It doesn't mean anything as I know. Yes it is meaningless. All these things doesn't oppose my argument in the question. – Sensebe Nov 21 '14 at 13:57
  • @AAA: Thank you for the comment. Can you say which comment I have skipped? – Sensebe Nov 21 '14 at 13:58
  • @Ant: Thank you for the involvement. I will for sure edit my question, if said, where the confusion in understanding going on. I really want to get this bothering thing out of my head. – Sensebe Nov 21 '14 at 14:06
  • Don't want to be rude. But how is it you doing vector calculus without understanding in straight forward calculus? Or is the vector part confusing? :/. – Chinny84 Nov 21 '14 at 14:07
  • @Chinny84: I am learning these all things (mathematical methods) in some other place other than college, specially meant for undergraduate student, to develop them for research at the early age. Sorry, I am not expert in calculus and vector part. I will learn if you can show the road to solving this problem for me. – Sensebe Nov 21 '14 at 14:12
  • I take it back then. But I do think you should try to start from non vector calculus and take it forward from there, because it could be a tricky road if you are trying to understand the more complicated topics without the prior knowledge to support you in your quest..good luck – Chinny84 Nov 21 '14 at 14:38
  • @Chinny84: Surely, I will understand, if you can answer the question I posed for Alan's comment above. – Sensebe Nov 21 '14 at 14:40
  • @Chinny84: I think I didn't make the question clear, I have edited it. Can you make some comment now? – Sensebe Nov 21 '14 at 15:18
  • That's not a real dot product in the usual sense. It's actually just a notational convention. – Akiva Weinberger Nov 21 '14 at 15:31
  • @columbus8myhw: Are you saying we are not performing dot product? – Sensebe Nov 21 '14 at 15:37
  • @Ant: I think my previous question was not clear, I have edited it. You clean up the unwanted comments. – Sensebe Nov 21 '14 at 15:38
  • @Godparticle Yes. Especially because $\frac\partial{\partial x}$ isn't a number and $\hat e_x\frac\partial{\partial x}$ isn't a vector. – Akiva Weinberger Nov 21 '14 at 15:39
  • @columbus8myhw: Yes, I agree with you. But, if we can't multiply those things, we shouldn't even differentiate, isn't it? – Sensebe Nov 21 '14 at 15:42
  • @Godparticle It's a notational convention, I think. Because it's so similar to the dot product, we use the same symbol, "$\cdot$". – Akiva Weinberger Nov 21 '14 at 15:47
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    Let us continue this discussion in chat. – Sensebe Nov 21 '14 at 15:54

4 Answers4

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

Umberto P.
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$(\frac{\partial }{\partial x})(x)$ 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.

Randy E
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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}

MBYagbasan
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  • Thank you for the answer. I think you didn't understand the question or I didn't understand you. I know that $\frac{\partial x}{\partial x}=1$, the question is by which multiplication rule we got $\frac{\partial x}{\partial x}$ in my question. – Sensebe Nov 21 '14 at 14:03
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    @Godparticle Not multiplication. It's application of an operator. – Daniel Fischer Nov 21 '14 at 14:07
  • In the problem I have given in the question, we are supposed to do multiplication. Isn't it? – Sensebe Nov 21 '14 at 14:13
  • it is not a multiplication. $\frac{\partial}{\partial x}$ is an operator. But the value of the operator at $x$ is $\frac{\partial x}{\partial x}$ can be seen as ratio of differential of $x$ to differential of $x$ so it is 1. But $\frac{\partial }{\partial x}$ is not a multiplication. It is derivation operator. – MBYagbasan Nov 21 '14 at 14:26
  • Yes, $\frac{\partial}{\partial x}$ is an operator. Yes, $\frac{\partial}{\partial x}$ is not the multiplication. We have this multiplication: $(\frac{\partial}{\partial x})\space(x)$, what is the answer for this multiplication operation? – Sensebe Nov 21 '14 at 14:32
  • Define an operator $T$ from spaces of functions which have continuous partial derivatives to spaces of continuous functions as follows: \begin{equation} T(f)=\frac{\partial f}{\partial x}\end{equation}. If you use $\frac{\partial}{\partial x}$ sign for this operator you get $\frac{\partial}{\partial x}(f)=\frac{\partial f}{\partial x}$. $\frac{\partial}{\partial x}(x)$ means the value of the operator $\frac{\partial}{\partial x}$ at the point $x$. – MBYagbasan Nov 21 '14 at 14:44
  • @MBYagbasan: Thank you for your time. I understand all those things, as I know. Ok, lets make it clear. While solving dot product, we need to multiply magnitudes of respective operands, isn't it? – Sensebe Nov 21 '14 at 14:48
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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.

MEDVIS
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  • "We can't write (eˆx∂∂x).(eˆxx) as the product of the magnitudes of (eˆx∂∂x) and (eˆxx) and the cosine of their angle". Aren't we performing dot product operation? – Sensebe Nov 21 '14 at 15:35
  • @Godparticle Yes we are. But what I'm saying is that you don't only deal with magnitudes and an angle. – MEDVIS Nov 21 '14 at 15:49
  • +1 If we are performing dot product, we should multiply the so called magnitudes (though we don't agree one of them as magnitude), isn't it? – Sensebe Nov 21 '14 at 15:52
  • @Godparticle Let me ask you something: Could you draw a vector whose magnitude is $\frac{\partial}{\partial x}$? – MEDVIS Nov 21 '14 at 16:18
  • Please answer my previous question before proceeding. – Sensebe Nov 21 '14 at 16:26
  • That's the answer. You can talk about magnitude of the vector, if that vector can be exist solely. How can you assign a magnitude to $\vec\nabla$ when it doesn't exist separately and always has a partner! – MEDVIS Nov 21 '14 at 16:32
  • You are welcome to the discussion: http://chat.stackexchange.com/rooms/18866/discussion-between-godparticle-and-columbus8myhw – Sensebe Nov 21 '14 at 16:33