# Conceptualisation of d/dx

1. Aug 1, 2012

### FlyingPanda

Hi all,

I was relatively comfortable with Leibniz's notation of dy, dx as representing infinitesimally small values of change in y and x, however I was watching a Math lecture yesterday where I saw something I was simply unable to conceptualise.

I've always seen the d as an operator, that functions in the same way as the sine, cosine, tan, log, etc functions. d(x) is taking the input x and performing the differential operation on it, outputting .dx -> which is the infinitesimally small change in the variable x. This by itself is meaningless, unless it is put in a statement with another infinitesimal, such as dy, whereby ratios give meaning to the expression.

However, I saw the use of the annihilation operator in the following expression:

(d/dx + x) y = dy/dx + xy

I know algebraically it works, but it treats d as if it is some simple constant. Seeing d/dx by itself was too difficult to conceptualise, I can't grasp the meaning of it. For me it was akin to seeing:

(sin( /sin(x) + x)y = sin(y)/sin(x) + xy.

So, my questions are:

1. Can you simply pull the input of a differential away from the expression
-> dx + x = (d + 1)x

2. How do you conceptualise what the meaning of d/dx is. What does it actually mean?

2. Aug 1, 2012

### pwsnafu

An operator whose input is a function and returns the derivative of that function with respect to the variable x.

3. Aug 1, 2012

### FlyingPanda

But how can an operator be added to 1, and then multiplied.

Let's take this case,

d(x^2)/dx + x^2 = 35

(in this case, x is 5)

(d/dx + 1)x^2 = 35

d/dx + 1 = 35/x^2

d/dx + 1 = 7/5

d/dx = 2/5

How can d/dx just simply be "given" a value?

4. Aug 1, 2012

### chiro

Hey FlyingPanda and welcome to the forums.

Are you familiar with a linear operator? Do you know what a matrix is?

The reason is that a differential operator is a linear operator, and understanding this can help in understand how this operator works. When you extend this idea to how you take say a function of a linear operator (as is done in operator algebra's and extended frameworks of linear algebra), this concept is solidified.

In fact if you ever see operator methods for solving DE's, you'll see that when you get things like 1/(D+4)f(x), you expand out the terms by doing what you do with polynomials which is long-division. This would seem un-reasonable, but when you look at the theory of how to calculate (functions of linear operators), then this will be easier to understand.

The short answer is that it's not easy if you try and think about it symbolically, but it becomes easier if you think about the differential operator being linear (think of the tangent nature of a derivative and what the derivative is: a slope or a linear rate of change at a particular point) and then if you want to get rigorous, you look at the theory of linear operators, which is something that John Von Neumann looked at.

5. Aug 1, 2012

### HallsofIvy

Because "1", in this case, is also an operator. It is the identity operator that takes every x to itself.

This last step is incorrect. You are NOT multiplying "d/dx+ 1" by x^2, you are applying the operator d/dx+ 1 to x^2. The error is exactly the same as if you were to say "sin x= 1 therefore sin= 1/x".

It can't. Your error is as I pointed out.