Visual derivative

  • Thread starter aquaregia
  • Start date
  • #1
19
0

Main Question or Discussion Point

I am learning calculus and I liked the idea of how limits are represented visually, as a removable discontinuity (a single point gap) in a function of x. When we started studying derivatives I immediately came to the conclusion that a derivative is like a single curve "gap" in a function of x and y, however after doing some searching I have not found a visual representation of the derivative being like that, and after graphing some 3-d functions they did not turn out like I expected.

For example the derivative of x^2 is: (x + dx)^2 - (x^2) / dx. Now replace the dx with y and you have (x + y)^2 - (x^2) / y, this should be a 3-d surface, right? And it should basically look like a plane with a slope of 2 with gap at y=0 since that is indeterminate because it would be 0/0, however the graph is a strange 3-d curve. Does any one know if my reasoning is wrong?
 
Last edited:

Answers and Replies

  • #2
441
0
For example the derivative of x^2 is: (x + dx)^2 - (x^2) / dx. Now replace the dx with y and you have (x + y)^2 - (x^2) / y,
What is your definition of a derivative?!?

y = x^2

then dy/dx = lim (as h approaches 0) ((f(x+h) - f(x))/h)
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,794
925
I am learning calculus and I liked the idea of how limits are represented visually, as a removable discontinuity (a single point gap) in a function of x. When we started studying derivatives I immediately came to the conclusion that a derivative is like a single curve "gap" in a function of x and y, however after doing some searching I have not found a visual representation of the derivative being like that, and after graphing some 3-d functions they did not turn out like I expected.

For example the derivative of x^2 is: (x + dx)^2 - (x^2) / dx. Now replace the dx with y and you have (x + y)^2 - (x^2) / y, this should be a 3-d surface, right? And it should basically look like a plane with a slope of 2 with gap at y=0 since that is indeterminate because it would be 0/0, however the graph is a strange 3-d curve. Does any one know if my reasoning is wrong?
You mean z= [(x+y)^2- x^2]/y? No, that is not a plane.
 
  • #4
19
0
Whatever, I figured it out. I just typed it into the graphing calculator wrong, but the basic idea is true.
 
  • #5
19
0
You mean z= [(x+y)^2- x^2]/y? No, that is not a plane.
Are you sure? I just graphed it, and it was, actually I graphed ((x+y)^2- x^2)/y, but I dont think switching the parenhesis style makes a difference. Of course its indeterminate at y = 0 because that is the deraritive.
 
Last edited:

Related Threads for: Visual derivative

  • Last Post
Replies
3
Views
2K
Replies
3
Views
1K
Replies
8
Views
3K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
Top