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Why df=(∂f/∂x)dx + (∂f/∂y)dy?

  1. Oct 14, 2011 #1
    let f(x,y)=0
    Why df=(∂f/∂x)dx + (∂f/∂y)dy?
     
  2. jcsd
  3. Oct 14, 2011 #2

    Bacle2

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    This is the definition of the total derivative, aka differential as I know it.

    df here gives you the equation of the tangent plane that approximates the change of the function near a point. Was that your question?
     
  4. Oct 14, 2011 #3

    HallsofIvy

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    If x and y are themselves functions of a parameter, say, t, then
    we can think of f(x, y)= f(x(t), y(t)) as a function of the single variable t and, by the chain rule:
    [tex]\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}[/tex]

    And then, the usual definition of the "differential" as df= (df/dt)dt gives
    [tex]df= \frac{\partial f}{\partial x}\frac{dx}{dt}dt+ \frac{\partial f}{\partial y}\frac{dy}{dt}dt= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y} dy[/tex]
     
    Last edited: Oct 15, 2011
  5. Oct 14, 2011 #4
    But how can I prove this [tex]\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}[/tex]?

    I know the basic operation in partial differentiation but i just not quite understand the theory behind it. Are there some proofs?
     
  6. Oct 15, 2011 #5
    A very informal (and possibly incorrect) proof I just thought of:

    [tex]df=df(x(t),y(t))=f(x(t+h),y(t+h))-f(x(t),y(t))=f(x(t+h),y(t+h))-f(x(t),y(t+h))+f(x(t),y(t+h))-f(x(t),y(t))=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy\Leftrightarrow \frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}[/tex]

    For something more formal http://math.uc.edu/~halpern/Calc.4/Handouts/Proofchainrule2dim.pdf
     
    Last edited by a moderator: Apr 26, 2017
  7. Oct 16, 2011 #6
    It's the chain rule, it's very used in calculus. You can see a demonstration in wikipedia.

    That can be deduced writing f(x,y) as Taylor's series (for multivariate functions), and going up to the 2nd term. To do it f only has to be differentiable 2 times in (a,b) neighbourhood.

    [tex]f(x,y) = f(a,b) + (\frac{\partial f}{\partial x}(a,b), \frac{\partial f}{\partial y}(a,b))\cdot (x-a, y-b)[/tex]

    Putting,
    [tex]x-a=\Delta x[/tex]
    [tex]y-b=\Delta y[/tex]
    [tex]f(x,y)-f(a,b)=\Delta f[/tex]

    When [tex]\Delta x \to 0[/tex] and [tex]\Delta y \to 0[/tex] you get that expression.
     
  8. Oct 16, 2011 #7

    Bacle2

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    Now put the two previous answers together (they did the hard work, I am just chiming-in), and do a δ-ε proof, showing that you can approximate the value of your function within ε>0 by using the right value of δ. This is for real-valued functions. If not, i.e., for Rn-valued maps , show:

    i)The differential df is a linear map:

    ii) ||f(x+h)-f(x)-hL(x)||/||h||→ 0

    as ||h||→0 is satisfied only by the differential L(x)=df

    In your case, you want to show that your function can be approximated to any degree of accuracy ε>0 by working within a ball B(x,δ), as all the other posters said.
     
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