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Implicit partial differentiation

  1. Jul 28, 2016 #1
    1. The problem statement, all variables and given/known data

    in the notes , 'by applying chain rule to LHS of the above equation ' , which equation is the author referring to ?
    it's given that
    f /x + (f/z)(z/x) = 0 ,
    As we can see , the function contain variable x , y and z

    2. Relevant equations


    3. The attempt at a solution
    why not
    f /x + (f/z)(z/x) +(f/y)(y/x) = 0 ???
     

    Attached Files:

  2. jcsd
  3. Jul 28, 2016 #2

    BvU

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    ##f(x,y,z) = 0##
     
  4. Jul 28, 2016 #3

    BvU

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    because that is not true. That is dividing, not differentiating.

    If you mean "why not$$
    {df\over dx} = f_x + f_y y_x + f_z z_x$$ then the answer is: that's exactly what the author is doing.
     
  5. Jul 28, 2016 #4

    BvU

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    Given up on the other thread ?
     
  6. Jul 28, 2016 #5

    BvU

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    Note: The author makes a notation mistake (he/she too !). It should have read:

    Differentiate (*) both sides wrt ##x##, $$
    {df\over dx} = {d\over dx} (0) $$​

    Do you see the difference ? Do you understand what is meant with a partial derivative ##\partial f\over\partial x## (often written as ##f_x##) as opposed to a total derivative ##df\over dx## ?
     
    Last edited: Jul 28, 2016
  7. Jul 28, 2016 #6

    BvU

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    First:
    not ##\quad f /x + (f/z)(z/x) = 0\quad ## but ##\quad f _x + f_z \;\displaystyle {\partial z\over\partial x} = 0##​
    Second:

    It's not given, but derived from ##z=f(x,y)##​
     
  8. Jul 28, 2016 #7
    no, i have attached the working in the previous thread, still waiting for your reply:smile:
     
  9. Jul 28, 2016 #8
    well , i couldnt understand why the author wrote ∂f/∂x + (∂f/∂z)(∂z/∂x) = 0 , i have no idea how he come up with this equation...In the notes, he said that byapplying chain rule to LHS of equation? which equation did he mean?
     
  10. Jul 28, 2016 #9

    Ray Vickson

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    You may be confused partly because the cited article writes confusingly. You can't have both ##f(x,y,z) = 0## and ##z = f(x,y)##: it is absolutely never allowed to use the same symbol ##``f"## to stand for two totally different functions in the same sentence or two. I am surprised the author would make such a blunder.

    Oh well, maybe that's just me being grumpy this morning.
     
  11. Jul 28, 2016 #10
    do you mean the equation should look like this ?
    $${df\over dx} = df/dx +d f/dy (dy/dx) + (df/dz)(d z/dx) $$

    why the LHS is
    ( df /dx ) shouldnt it = 0 ? , just like the author's working ?
     
  12. Jul 28, 2016 #11

    BvU

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    We'll humour mfb and write$$f(x,y,z) = f\left (x,y,z\left (x,y\right )\right ) = 0$$ Apply the definition of derivative $$ {dg\over dx} \equiv\ \ \lim\limits_{h\downarrow 0} {g(x+h)-g(x)\over h}$$while y, as the other independent variable, is kept constant. Since ##f=0## and stays 0, this derivative should give 0.

    this is how the author was lured into writing ##\displaystyle {\partial f\over\partial x}##
    and I don't know how to do it didactically foolproof either. The point is that y is kept constant (because it is an independent variable) but z is a function of x so z does vary -- and thereby influences the derivative of f wrt x.​

    So what we get is $$ 0 =
    \ \ \lim\limits_{h\downarrow 0} {f(x+h, y, z(x+h,y)) - f(x,y,z(x,y))\over h} = \\
    \quad \lim\limits_{h\downarrow 0} {f(x+h, y, z(x+h,y)) - f(x,y,z(x+h,y) )\over h} + {f(x, y, z(x+h,y)) - f(x,y,z(x,y) )\over h}
    $$
    First term is ##\displaystyle {\partial f\over\partial x}## a.k.a. ##f_x##

    Second term is ##\displaystyle {\partial f\over\partial z} \displaystyle {\partial z\over\partial x}## a.k.a. ##f_z z_x##
    In summary: $$f_x + f_z z_x = 0$$

    Look up chain rule in your textbook.
     
  13. Jul 28, 2016 #12

    BvU

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    quoting a quote looks messy in PF. Can you please distinguish between ##d## and ##\partial##, otherwise we might as well type chinese characters.

    The author confuses you ( and irritates mfb and me also) by using the same ##\partial f\over \partial x## for two different things:

    If he/she writes ##{\partial f\over \partial x} = {\partial \over \partial x} (0) ## (which is 0)
    that means: while keeping y the same and letting z vary with x
    (it is a partial derivative because f is a function of two independent variables).​

    If he/she writes ## {\partial f\over \partial x} + {\partial f\over \partial z} {\partial z\over \partial x } =0 ##
    that means: while keeping y and z the same. Now f is considered as a function of three ' independent' variables.
    Drives one crazy, I agree. Still, it's elementary.
     
  14. Jul 28, 2016 #13

    BvU

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    Oh, missed the alert. Will look.

    Nope, you missed the alert: I posted #67 with more 'suggestions for improvement' :smile:
     
  15. Jul 28, 2016 #14
    can you explain the working? why we need to break the working into 2 parts?
     

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  16. Jul 28, 2016 #15
    well you said that we keep y and z same now....Why there's (∂f/∂z)(∂z/∂x)appear in equation....
    (∂f/∂z)(∂z/∂x) mean function f(x,y,z) varies with z,am i right?
     
  17. Aug 10, 2016 #16

    BvU

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    To show you how the two terms of ##\displaystyle {\partial f\over \partial x} ##, namely ##\ f_x\ ## and ##\ f_z \displaystyle {\partial z \over \partial x} \ ## come about.
     
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