Using Chain Rule to Find du/dT & du/dv: Step-by-Step Guide

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This discussion focuses on applying the chain rule to find the derivatives du/dT and du/dv for a function u = u(T, v). The chain rule is expressed as du/dx = (∂u/∂T)(dT/dx) + (∂u/∂v)(dv/dx), where T and v are functions of x. The distinction between the symbols ∂ (partial derivative) and d (total derivative) is clarified, emphasizing that ∂ is used in multivariable calculus to denote derivatives while treating other variables as constants. An example is provided to illustrate the calculation of partial derivatives.

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If a function is given by u = u(T,v) how to use the chain rule to write how u changes with respect to T & v.
Please specify the steps involved.
i understand chain rule as \frac{du}{dx} = \frac{du}{dy} \frac{dy}{dx}
 
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Assuming T and v are both functions of x, the chain rule gives:
du/dx = (∂u/∂T)(dT/dx) + (∂u/∂v)(dv/dx)
 
mathman said:
Assuming T and v are both functions of x, the chain rule gives:
du/dx = (∂u/∂T)(dT/dx) + (∂u/∂v)(dv/dx)
very much thanks for ur quick response :)

how does the symbol differs in meaning with d
 
\partial (read "partial") is basically the equivalent of d in multivariable calculus, it means you take the derivative of a function with respect to a variable while considering all other variables as constants during that operation.

For example if I have
F = x^2 + 2xy + \frac{x}{y}
Then
\frac{\partial F}{\partial x} = 2x + 2y + \frac{1}{y}
Can you find \frac{\partial F}{\partial y}? :)
 
JHamm said:
<br /> Can you find \frac{\partial F}{\partial y}? :)
<br /> <br /> = 2x - x/y<sup>2</sup><br /> <br /> i think that&#039;s the right answer :)
 
I think so too :)
 
Appreciate ur help :)
 

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