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Proof of Multivariable chain rule

  1. Nov 7, 2011 #1
    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.
  2. jcsd
  3. Nov 7, 2011 #2


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    Science Advisor

    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.
  4. Nov 7, 2011 #3
    The MIT OCW videos on multivariable calculus have video which covers this: http://ocw.mit.edu/courses/mathemat...fall-2007/video-lectures/lecture-12-gradient/

    To summarize the argument (though I doubt this is particularly rigorous)
    df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz
    Then, if we assume that f, x, y and z are all functions of t we divide by dt
    \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}
    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.
  5. Nov 7, 2011 #4


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    Homework Helper

    Same proof as the single variable chain rule.
  6. Nov 8, 2011 #5
    Makes sense. thanks.
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