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Partial derivative chain rule

  1. Oct 13, 2013 #1
    could someone show me how [itex]\frac{∂}{∂t}[/itex]([itex]\frac{∂z}{∂u}[/itex])= [itex]\frac{∂^2z}{∂u^2}[/itex] [itex]\frac{∂u}{∂t}[/itex]

    where z=z(u)

    u=x+at
     
  2. jcsd
  3. Oct 13, 2013 #2

    LCKurtz

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    ##z## is only a function of ##u## so you might as well use ##z'(u)## instead of partials. I will denote ##\frac{\partial z}{\partial u}## as ##z'(u)##. Your equation then becomes$$
    \frac \partial {\partial t}z'(u) = \frac {d}{du}z'(u)\frac{\partial u}{\partial t}
    =z''(u)\frac{\partial u}{\partial t}$$Isn't that exactly the chain rule on ##z'(u)##?

    [Edit] Corrected some TeX partials vs d/dt symbols.
     
    Last edited: Oct 13, 2013
  4. Oct 13, 2013 #3
    i apologize, but can you please show me what you did step by step? i know how to apply the chain rule for functions when the function itself is actually given, ie the derivative of the function times the derivation of the inside. but i don't understand what we're doing now.

    where did you get the d/du in the front from? and the ∂u/∂t that's behind z', that's just from the derivative of the inside right?
     
  5. Oct 13, 2013 #4

    LCKurtz

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    If ##f## is a function of ##u## and ##u## is a function of ##x## and ##y## then ##f## is ultimately a function of ##x## and ##y##. It makes sense to ask for the partials of ##f## with respect to ##x## and ##y##. The corresponding chain rules are (I will use subscripts for partials to save typing): $$
    f_x = f'(u)u_x,~f_y = f'(u)u_y$$Note the use of the ' on ##f## since ##f## itself is a function of just one variable and the partial symbols on ##u## since it is a function of more than one variable. These formulas are almost certainly in your text.

    Your problem is solved by applying one of these chain rules to ##z'(u)## with second variables ##x## and ##t## instead of ##x## and ##y##.
     
  6. Oct 13, 2013 #5
    i understand z'(t)=[itex]\frac{∂z}{∂u}[/itex][itex]\frac{∂u}{∂t}[/itex] and z'(x)=[itex]\frac{∂z}{∂u}[/itex][itex]\frac{∂u}{∂x}[/itex]


    i just don't understand the leap from that to this: [itex]\frac{∂}{∂t}[/itex][itex]\frac{∂z}{∂u}[/itex]

    but 'u' is the intermediate variable. i don't know what to do with [itex]\frac{∂z}{∂u}[/itex]. if it involved finding the partial of z with respect to x and t, i could just apply the chain rule like you say, but it's asking me to find the partial of z with respect to u, the intermediate variable. what can i do with that? i can't use x and y.
     
    Last edited: Oct 13, 2013
  7. Oct 13, 2013 #6

    LCKurtz

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    It becomes exactly what I showed you in post #2.
     
  8. Oct 13, 2013 #7
    okay can you tell me if i did this correctly

    [itex]\frac{∂}{∂t}[/itex]([itex]\frac{∂z}{∂u}[/itex])=[itex]\frac{∂}{∂u}[/itex]([itex]\frac{∂z}{∂t}[/itex])=[itex]\frac{∂}{∂u}[/itex]([itex]\frac{∂z}{∂u}[/itex][itex]\frac{∂u}{∂t}[/itex])

    i basically just swapped the positions of the u and t in the beginning is this right?

    if i swap the u and the t, aren't i applying clairaut's theorem? and if i am, then don't i need to prove that zut is continuous and ztu is continuous since that's the condition for which clairaut's theorem applies?
     
  9. Oct 14, 2013 #8

    LCKurtz

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    That is not the setting for Clairaut's theorem. I don't know why you are doing this. The answer to your question in your post #1 is that it is the chain rule I explained in post #4. That is where the formula comes from, and the proof of the chain rule is probably in your text.
     
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