Multivariable calculus, partial derivatives

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SUMMARY

This discussion focuses on the application of the chain rule in multivariable calculus, specifically in the context of partial derivatives. Participants clarify the notation and concepts surrounding the composition of functions, denoted as ##F \circ G##, where ##G(s,t) = (u(s,t), v(s,t))##. The chain rule is emphasized as a method to compute derivatives of composite functions, leading to the expression for ##\frac{\partial W}{\partial s}##. The conversation highlights the confusion around notation and the necessity of understanding the underlying functions involved.

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  • Understanding of multivariable calculus concepts
  • Familiarity with the chain rule for derivatives
  • Knowledge of function composition in calculus
  • Ability to interpret mathematical notation and expressions
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  • Study the chain rule in multivariable calculus in detail
  • Practice problems involving partial derivatives and function composition
  • Learn about the notation used in multivariable calculus
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Students and educators in mathematics, particularly those studying or teaching multivariable calculus, as well as anyone seeking to improve their understanding of partial derivatives and the chain rule.

Feodalherren
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Homework Statement


Untitled.png



Homework Equations





The Attempt at a Solution


Umm can somebody explain to me what just happened. None of that makes any sense to me what so ever.
 
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Are you familiar with the chain rule?
 
Of course. The notation in this problem is so confusing that I can't follow what's happening.
 
Let's use the notation ##G(s,t) = (u(s,t), v(s,t))##.

What you need to do is to compute

\frac{\partial}{\partial s} F\circ G

What will that be according to the chain rule?
 
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Correct?
 
What happened to ##F##? I can only see ##G## showing up.
 
Ok I have no idea.. I can't remember what FoG means :/. You didn't even define F as anything?
 
Feodalherren said:
Ok I have no idea.. I can't remember what FoG means :/. You didn't even define F as anything?

##F## is just an arbitrary differentiable map (as shown in the problem statement).

What ##F\circ G## means is that it is the map which sends ##(s,t)## to ##F(G(s,t))##.

Can you show me the chain rule you've learned (and perhaps the variations)?
 
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I've only learned the one that I demonstrated. You draw one of those trees and then just differentiate however many times you need to in order to get to the variable that you need.

I'm totally lost on the notation yet again.
 
  • #10
Feodalherren said:

Homework Statement


Untitled.png



Homework Equations





The Attempt at a Solution


Umm can somebody explain to me what just happened. None of that makes any sense to me what so ever.

Is the "subscript" notation throwing you off? If so, forget it and use a more exact nomenclature: ##W = F(u,v)## gives
\frac{\partial W}{\partial s} = \frac{\partial F}{\partial u} \frac{\partial u}{\partial s}<br /> + \frac{\partial F}{\partial v} \frac{\partial v}{\partial s}\: \longleftarrow \text{ chain rule}\\ <br /> \text{ }\\ \text{or}\\ <br /> \text{ }\\<br /> W_s = F_u u_s + F_v v_s
etc.
 
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  • #11
Ok that helps Ray. Now for the evaluation.

If I know Ws (1,0) then
s=1, t=0

Then I get:
u (1,0) and s (1,0).

Where do I evaluate the first therm dF/du?
 

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