# Visual derivative

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?

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## Answers and Replies

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)

HallsofIvy
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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.

Whatever, I figured it out. I just typed it into the graphing calculator wrong, but the basic idea is true.

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.

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