Prove the equality : Multivariable chain rule problem

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SUMMARY

The discussion centers on proving the equality \((\frac{\partial u}{\partial x})^{2} + (\frac{\partial u}{\partial t})^{2} = e^{-2s}[(\frac{\partial u}{\partial s})^{2} + (\frac{\partial u}{\partial t})^{2}]\) using the multivariable chain rule. The variables are defined as \(u = f(x,y)\), \(x = e^{s}\cos(t)\), and \(y = e^{s}\sin(t)\). The initial approach involved computing the partial derivatives \(\frac{\partial u}{\partial s}\), \(\frac{\partial u}{\partial x}\), and \(\frac{\partial u}{\partial y}\), but the user encountered complexity. A suggestion was made to evaluate the right side directly, which simplifies the process by eliminating unnecessary terms.

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  • Experience with implicit differentiation techniques.
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  • Practice calculating partial derivatives for functions of multiple variables.
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Students and educators in calculus, particularly those focusing on multivariable functions and the chain rule, as well as anyone seeking to deepen their understanding of partial derivatives and their applications in mathematical proofs.

michonamona
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Homework Statement


Prove that

[tex](\frac{\partial u}{\partial x})^{2} + (\frac{\partial u}{\partial t})^{2} = e^{-2s}[(\frac{\partial u}{\partial s})^{2} + (\frac{\partial u}{\partial t})^{2}].[/tex]

Homework Equations



[tex]u = f(x,y)[/tex]
[tex]x = e^{s}cost[/tex]
[tex]y = e^{s}sint[/tex]

The Attempt at a Solution



I started out by computing [tex]\frac{\partial u}{\partial s}[/tex], then solving it for [tex]\frac{\partial u}{\partial x}[/tex]. Then I did the same for [tex]\frac{\partial u}{\partial y}[/tex]. So I got some messy equations, that made think that there must be a much easier way to solve this. I also tried implicit differentiation but got stuck. Any insight?

Thanks,
M
 
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Start by evaluating the right side, and don't try to solve for ux or uy. If you calculate the two partials on the right side correctly, many terms will drop out.
 
Ah...thank you.
 

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