Proof of Multivariable chain rule

  • #1
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Main Question or Discussion Point

I was wondering how to prove the multivariable chain rule

[tex]\frac{\mbox{d}z}{\mbox{d}t}=\frac{\partial z}{\partial y}\frac{\mbox{d}y}{\mbox{d}t}+\frac{\partial z}{\partial x}\frac{\mbox{d}x}{\mbox{d}t}[/tex]

where [tex]z=z(x(t),y(t))[/tex]

I don't really need an extremely rigorous proof, but a slightly intuitive proof would do.

Also how does one prove that if z is continuous, then

[tex]\frac{{\partial}^{2}z}{\partial x \partial y}=\frac{{\partial}^{2}z}{\partial y \partial x}[/tex]

Thanks in advance.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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As for your second question, one doesn't- what you have written is not true. If z is only continuous, the partial derivative, much less the second derivatives, may not even exist. What you need is that the second derivatives are continuous.
 
  • #3
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The MIT OCW videos on multivariable calculus have video which covers this: http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-12-gradient/

To summarize the argument (though I doubt this is particularly rigorous)
[tex]
df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz
[/tex]
Then, if we assume that f, x, y and z are all functions of t we divide by dt
[tex]
\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}
[/tex]
Again this is not rigorous, if you want an idea of why this could be true think of the way a vector is composed by summing orthogonal parts; take the amount f will change when you move an amount along the x direction, multiplied by the amount x will move when you change t a certain amount, then repeat this for y and z and sum for the final change in f.
 
  • #4
lurflurf
Homework Helper
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Same proof as the single variable chain rule.
 
  • #5
371
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The MIT OCW videos on multivariable calculus have video which covers this: http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-12-gradient/

To summarize the argument (though I doubt this is particularly rigorous)
[tex]
df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz
[/tex]
Then, if we assume that f, x, y and z are all functions of t we divide by dt
[tex]
\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}
[/tex]
Again this is not rigorous, if you want an idea of why this could be true think of the way a vector is composed by summing orthogonal parts; take the amount f will change when you move an amount along the x direction, multiplied by the amount x will move when you change t a certain amount, then repeat this for y and z and sum for the final change in f.
Makes sense. thanks.
 

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