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Chain rule for partial derivatives

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data


    So there is an exercise in which I should "verify" the chain rule for some functions.
    In other words to do it by substitution, then doing by the formula and checking if the results are the same. (and checking with the book`s answer too)

    For a few of them, they just don`t match.
    for all the rest they match.

    I'm really not sure what exactly I`m doing wrong .
    So I want to check my method.

    I'll just post one of them, if it really is necessary I'll post another one.


    2. Relevant equations

    chain rule: [itex] \frac{\mathrm{d} h}{\mathrm{d} t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}
    [/itex]


    1st function: [itex]f(x,y)= \ln (x^2 + y^2)
    [/itex]
    where: [itex]x = 2t + 1[/itex] and [itex]y = 4t^2 -5 [/itex]

    3. The attempt at a solution

    Substitution

    [itex]h(t) = \ln ((2t + 1)^2 + (4t^2 - 5)^2 )
    [/itex]

    then: [itex]\frac{\mathrm{d} h}{\mathrm{d} t} = \frac{1}{((2t+1)^2 + (4t^2-5)^2)}\cdot [ (2)(2)(2t+1)+ (2)(8t)(4t^2 - 5)][/itex]

    which is: [itex]\frac{\mathrm{d} h}{\mathrm{d} t} = \frac{(8t+4)+(64t^3-80t)}{((2t+1)^2 + (4t^2-5)^2)} [/itex]

    which is [itex]\frac{\mathrm{d} h}{\mathrm{d} t} = \frac{64t^3 - 72t +4 }{16t^4 -36t^2 +2t + 26}[/itex]

    so it is [itex]\frac{\mathrm{d} h}{\mathrm{d} t} = \frac{32t^3 -36t + 2}{8t^4 - 18t^2 + t +13 }[/itex]


    now by the chain rule formula

    [itex]
    \frac{\mathrm{d} h}{\mathrm{d} t} = (\frac{2x}{x^2 + y^2}\cdot 2t) + (\frac{2y}{ x^2+y^2}\cdot 8t)[/itex]


    [itex] \frac{\mathrm{d} h}{\mathrm{d} t} = \frac{2(2t + 1)2t}{(2t+1)^2 + (4t^2-5)^2} + \frac{2(4t^2-5)8t}{ (2t+1)^2+(4t^2-5)^2} [/itex]

    [itex]
    \frac{\mathrm{d} h}{\mathrm{d} t} = \frac{8t^2 + 4t }{(2t+1)^2 + (4t^2-5)^2}+ \frac{64t^3 - 80t }{ (2t+1)^2+(4t^2-5)^2} = \frac{64t^3 +8t^2 - 76t}{16t^4-16t^2 +4t + 26 } [/itex]


    [itex]
    \frac{\mathrm{d} h}{\mathrm{d} t}= \frac{36t^3 +4t^2 -38}{8t^4 - 8t^2 +2t + 13}
    [/itex]


    Which not only it is different from the first method but it is also different form the answer in the book. which is

    [itex] \frac{32t^3 - 36t + 2}{8t^4 - 18t^2 + 2t + 13 }
    [/itex]


    edit: looking now the first method is almost right except for that 2t...
     
  2. jcsd
  3. Nov 8, 2011 #2

    lurflurf

    User Avatar
    Homework Helper

    The book answer is right. There are a few errors in your workings, one of which is in the chain rule
    [itex]
    \frac{\mathrm{d} h}{\mathrm{d} t} = (\frac{2x}{x^2 + y^2}\cdot 2t) + (\frac{2y}{ x^2+y^2}\cdot 8t)[/itex]
    should be
    [itex]
    \frac{\mathrm{d} h}{\mathrm{d} t} = (\frac{2x}{x^2 + y^2}\cdot 2) + (\frac{2y}{ x^2+y^2}\cdot 8t)[/itex]
     
  4. Nov 10, 2011 #3
    Okay.

    This time I did it more carefully and it was correct.
    Did it for the other few ones that weren't matching and now they match too.

    basically, I just need to do it more carefully.


    Thanks for helping!
     
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