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Multidimensional chain rule (tough)

  1. May 19, 2014 #1

    joshmccraney

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    Gold Member

    hey pf!

    suppose i have a function ##f( x , y)##. i make a change of variables such that ##z(x,y)## in such a way that now ##f( z , y)##. how do i find $$\frac{\partial f}{\partial y}$$ $$\frac{\partial f}{\partial x}$$ $$\frac{\partial^2 f}{\partial y^2}$$ $$\frac{\partial^2 f}{\partial x}$$

    i think $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial z}\frac{\partial z}{\partial x}$$ and $$\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 f}{\partial z^2} \frac{\partial z}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial^2 z}{\partial x^2}$$

    i also think $$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial z} \frac{\partial z}{\partial y} + \frac{\partial f}{\partial y}$$ but something is wrong here. i feel that i need some new notation or something to fully represent what is happening.

    i have no idea how to express $$\frac{\partial^2 f}{\partial y^2}$$

    any help is greatly appreciated
     
  2. jcsd
  3. May 20, 2014 #2
    Great inquiry and along your investigation you even noted the vulgarity of the notation!

    To free yourself from the corruption the present notation is proving, I recommend the following:
    you started from a function of x and y and then switched to new coordinates x' and y' by the transformation

    x'=z(x,y)
    y'=y

    This cleanses the notation. Redo your analysis where derivatives are dressed either primed or unprimed. You will see that your results are correct so long as you change some of their form.

    (Hint: write down the f as a function of the unprimed coordinates and a take a partial of some primed coordinate, say x'. Carry out chain rules.)
     
    Last edited: May 20, 2014
  4. May 20, 2014 #3

    HallsofIvy

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    The notation confusion comes where you chose to use "y" as both a variable in the first function and a variable in the definition of "x". If you are going to have two variables it would be better to use two new symbols.

    That is, you initially have f(x, y) and you have a new variable, z, that is, in some way, a function of both x and y so you write f(z, y). That notation makes no sense to me. If this new z is a function of x and y wouldn't it make equal sense to say "f(x, z)"? The difficulty is that you want to replace x or y with a new variable so that instead of knowing that z is a function of x and y, you need to know how x, and/or y, is a function of z.

    For example, if we have f(x,y)= x+ y and are given z= x/y, we can write x= yz and so can write f(z, y)= yz+ z= z(y+ 1). But we could as well write y= x/z and write f(x, z)= x+ (x/z)= x(z+ 1)/z.
     
  5. May 21, 2014 #4

    joshmccraney

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    Gold Member

    Hey thanks to you both!!!!!
     
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