How do partial derivatives relate to the definition of a derivative?

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Homework Help Overview

The discussion revolves around the concept of partial derivatives in the context of multivariable calculus, specifically how they relate to the definition of derivatives. The original poster explores the geometric interpretation of partial derivatives and their calculation using a specific function.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions the geometric meaning of partial derivatives and why only one variable is varied while others are held constant. They also seek clarification on their calculation process and the correct notation for derivatives.

Discussion Status

Participants are providing feedback on the original poster's understanding and calculations. Some have pointed out errors in differentiation and emphasized the importance of proper notation. There is an ongoing exploration of the rationale behind keeping other variables constant when calculating partial derivatives.

Contextual Notes

Participants mention practical applications of partial derivatives in various fields, suggesting that understanding their behavior is crucial for analyzing multivariable functions.

courtrigrad
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I am sort of skipping around at my own pace in Courant's Calculus book and came across partial derivatives. Are they geometrically the intersection of a plane and a surface? Why do we keep only one variable changing and the other variables fixed? Is it basically the definition of the derivative? For example

[tex]\frac{\partial y}{\partial x}[/tex] when [tex]f(x,y) = \sqrt{x^{2} + y^{2}}[/tex] Ok so would I consider y to be a constant when we want to find [tex]f_{x}[/tex] and vice versa for [tex]f_{y}[/tex]? Ok so this is what I did:

[tex]f_{x} = \frac{1}{2}\sqrt{x^2+y^2}2x[/tex]. But the answer is:

[tex]f_{x} = \frac{x}{\sqrt{x^2+y^2}}[/tex] and the same is true for [tex]f_{y} = \frac{y}{\sqrt{x^2+y^2}}[/tex] except the variables are reversed.

Any help is appreciated!

Thanks
 
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Did you mean df/dx and not dy/dx? When you evaluate the partial derivative wrt a particular variable, you keep the others constant as you said.

Your answer is wrong as you've not differentiated with the chain rule properly:

d/dx [(x^2 + y^2)^1/2)] = (1/2)[(x^2 + y^2)^(-1/2)].2x = x/(x^2 + y^2)^1/2 as required.
 
whoops I must have not noticed that I typed LaTex wrong.

Thanks a lot for your answer :smile:
 
Please use the notation of Lagrange properly.
[tex]f'_{x}=:\frac{\partial f}{\partial x}[/tex]

,where the last is C.G.Jacobi's notation.

Daniel.
 
courtrigrad said:
Why do we keep only one variable changing and the other variables fixed?
I'm sure you can think of many multivariable functions where you are only interested in what happens when one particular variable is varied (Ie., gas laws, economics, etc.). In addition, partials are useful in general form as they make studying the derivative, and thus properties of a function easier, as the derivative can be written in terms of the partial derivatives of f.
 
courtigrad:
The simplest way of looking upon a partial derivative of a function f, is that it measures the rate of change of f along a RESTRICTED neigbourhood of your evaluation point.
That is, [tex]\frac{\partial{f}}{\partial{x}}\mid_{(\vec{x}=(x_{0},y_{0}))[/tex] is found by by restricting your attention to f's behaviour along the line [tex]y=y_{0}[/tex] (where "y" is obviously a constant!)

The one variable analogue of the partial derivative, is to limit your attention to f's behaviour on a resticted neighbourhood (for example, by evaluating the rate of change on the rational sequences converging to your point, not bothering about f's behaviour on sequences converging to your point where the elements of the sequences are irrational).
 
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