Conceptualisation of d/dx

  • Thread starter FlyingPanda
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In summary: 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".
  • #1
FlyingPanda
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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?

Thanks for your help!
 
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  • #2
FlyingPanda said:
2. How do you conceptualise what the meaning of d/dx is. What does it actually mean?

An operator whose input is a function and returns the derivative of that function with respect to the variable x.
 
  • #3
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
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
FlyingPanda said:
But how can an operator be added to 1, and then multiplied.
Because "1", in this case, is also an operator. It is the identity operator that takes every x to itself.

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
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".

d/dx + 1 = 7/5

d/dx = 2/5

How can d/dx just simply be "given" a value?
It can't. Your error is as I pointed out.
 

1. What is the concept of d/dx in calculus?

The concept of d/dx in calculus is known as the derivative or the rate of change of a function with respect to its independent variable. It is represented as dy/dx and measures the instantaneous slope of a curve at a specific point.

2. How is d/dx calculated?

The derivative dy/dx is calculated using the limit definition of the derivative, which involves taking the limit as the change in x approaches 0. Alternatively, it can be calculated using rules such as the power rule, product rule, quotient rule, and chain rule.

3. What is the geometric interpretation of d/dx?

The geometric interpretation of d/dx is the slope of the tangent line of a curve at a specific point. It represents the rate of change of a function at that point and can be negative, positive, or zero depending on the direction of the curve.

4. Why is d/dx important?

The concept of d/dx is important in calculus as it allows us to analyze the behavior and properties of functions. It is used to find the maximum and minimum values of a function, determine the concavity of a curve, and solve optimization problems.

5. How is d/dx used in real-world applications?

In real-world applications, d/dx is used to model and analyze various phenomena such as motion, growth, and decay. It is used in fields such as physics, engineering, economics, and statistics to make predictions and optimize systems.

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