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Question on derivative of multi variable.

  1. Feb 6, 2010 #1
    I have a hard time taking the derivative [tex] \frac{\partial u}{\partial A} [/tex]

    Where [tex] u=A+B[/tex] and [tex]A = x+2y,B = x-2y [/tex]

    This is because A and B are both function of x and y. I don't think I can treat A and B as totally indenpend to each other where

    [tex] \frac{\partial B}{\partial A}=0, \frac{\partial A}{\partial B}=0 [/tex] .

    This is what I did and please tell me what did I do wrong.

    [tex]A = x+2y,B = x-2y \Rightarrow x=\frac{A +B}{2}, y=\frac{A - B}{4} [/tex]

    [tex]\Rightarrow \frac{\partial A}{\partial x}= 1 , \frac{\partial A}{\partial y}= 2 , \frac{\partial B}{\partial x}= 1 , \frac{\partial B}{\partial y}= -2[/tex]

    [tex]x=\frac{A +B}{2}\Rightarrow \frac{\partial x}{\partial A}=\frac{1}{2} [\frac{\partial A}{\partial A}+\frac{\partial B}{\partial x}\frac{\partial x}{\partial A}] = \frac{1}{2} [1 + \frac{\partial B}{\partial x}\frac{\partial x}{\partial A}] = \frac{1}{2} + \frac{1}{2}\frac{\partial x}{\partial A}\Rightarrow \frac{\partial x}{\partial A} = 1[/tex]

    [tex]y=\frac{A - B}{4} \Rightarrow \frac{\partial y}{\partial A}=\frac{1}{4} [\frac{\partial A}{\partial A}- \frac{\partial B}{\partial y}\frac{\partial y}{\partial A}] = \frac{1}{4} [1 + 2\frac{\partial y}{\partial A}] = \frac{1}{4} + \frac{1}{2}\frac{\partial y}{\partial A}\Rightarrow \frac{\partial y}{\partial A} = \frac{1}{2}[/tex]

    [tex]\frac{\partial y}{\partial B}=\frac{1}{4} [\frac{\partial A}{\partial y}\frac{\partial y}{\partial B} - \frac{\partial B}{\partial B}] = \frac{1}{4} [ 2\frac{\partial y}{\partial B} - 1] = \frac{1}{2}\frac{\partial y}{\partial B} -\frac{1}{4} \Rightarrow \frac{\partial y}{\partial B} = -\frac{1}{2}[/tex]

    [tex]\frac{\partial u}{\partial A} = \frac{\partial A}{\partial A}+ \frac{\partial B}{\partial x}\frac{\partial x}{\partial A} + \frac{\partial B}{\partial y}\frac{\partial y}{\partial A} = 1 +\frac{\partial B}{\partial x} + \frac{1}{2}\frac{\partial B}{\partial y} = 0[/tex]

    Obviously I did it wrong, can anyone tell me what I did wrong. This is not homework. Question is how to treat A and B when they both are function of x & y.

    Last edited: Feb 7, 2010
  2. jcsd
  3. Feb 7, 2010 #2


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    So u= (x+2y)+ (x- 2y)= 2x?
    [tex]\frac{\partial u}{\partial x}= 2[/tex]
    [tex]\frac{\partial u}{\partial y}= 0[/tex]

    Or the chain rule:
    [tex]\frac{\partial u}{\partial x}= \frac{\partial u}{\partial A}\frac{\partial A}{\partial x}+ \frac{\partial u}{\partial B}\frac{\partial B}{\partial x}[/tex]
    [tex]= 1(1)+ 1(1)= 2[/tex]

    [tex]\frac{\partial u}{\partial y}= \frac{\partial u}{\partial A}\frac{\partial A}{\partial y}+ \frac{\partial u}{\partial B}\frac{\partial B}{\partial y}[/tex]
    [tex]= 1(2)+ 1(-2)= 0[/tex]

    That doesn't matter.

    I don't know that there is any thing wrong with what you did, but I don't know why you would do it. Neither the derivative of x with respect A nor the derivative of y with respect to B is relevant to your original question.
  4. Feb 7, 2010 #3
    Thanks for your reply.

    I gave a bad example that it simplify into 2x. I since read the Chain Rule again. I just want to double check with you:

    Assumming [tex] A=x+2y, B=cos(x)+sin(y)[/tex] so it cannot be simplify like the other one. Please tell me the two points I make below is correct:

    1) My understanding now is in order for [tex]\frac{\partial u}{\partial A}[/tex] to be valid, A has to be a function of u. [tex]\frac{\partial A}{\partial u}[/tex] is invalid because u is not a function of A and therefore [tex]\frac{\partial A}{\partial u} = 0[/tex].

    2) THis also true that A is not a function of B or the other way around EVEN both contain the same two variable x and y. Therefore
    [tex]\frac{\partial A}{\partial B} = 0,\frac{\partial B}{\partial A} = 0 [/tex].

    Last edited: Feb 7, 2010
  5. Feb 7, 2010 #4


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    You probably meant to say "because A is not a function of u". But if u=A+B, then A=u-B. So A does depend on u. (u,A, and B all depend on each other, namely because they all depend on x and y and those are the actual variables)
  6. Feb 7, 2010 #5
    Can you tell me how should I look at it? I am still pretty confused.
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