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Partial derivatives + Chain rule

  1. Jan 28, 2004 #1
    I am having a hard time doing the following problems. First off all the notation is confusing the hell out of me. This is the first time i have used this notation so it is making learning very difficult. Here are my questions.

    Prove the following function is differentiable, and find the derivative at an arbitarary point.

    1.) f:R2-->R,(x,y)|--> exy

    2.) let f(u,v,w) = (eu-w, cos(v+u) + sin(u+v+w)) and
    g(x,y) = (ex, cos(y-x), e-y). Calculate fog and D(fog)(0,0).

    Thanks.
     
  2. jcsd
  3. Jan 28, 2004 #2

    Tom Mattson

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    OK, first for the notation. It says is that your function is a mapping of R2 onto R1. Operationally, all that means is that you are accepting 2 independent variables to determine a single dependent variable.

    Second, I don't know why it says to calculate "the" derivative, because this thing has two first partials.

    Third, to prove differentiability, you have to prove that the limits:

    ∂f/∂x=limΔx-->0{f(x+Δx,y)-f(x,y)}/Δx
    ∂f/∂y=limΔy-->0{f(x,y+Δy)-f(x,y)}/Δy

    exist.

    Fourth, taking the derivative is a piece of cake. When taking the partial with respect to x (or y), just treat it as the ordinary derivative, and treat the y (or x) as a constant.

    Why is there a comma between the exponential and cosine functions?

    Again, why the commas?

    I think you'll have to be more specific with this notation, because "fog" refers to the composite function f(g(x,y)), in which g(x) is substituted for the independent variable in f(u,v,w). But your f(u,v,w) does not have just one independent variable, but three. There is ambiguity in specifying "fog".

    So, let me ask you:

    For precisely which independent variable of f(u,v,w) are you to substitute g(x,y)?

    edit: typo
     
  4. Jan 28, 2004 #3
    as for fog that means the composite function. And as for the commas that is how the book has it.
     
  5. Jan 29, 2004 #4

    matt grime

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    for (x,y) maps to e^(xy)

    note that for fixed y, the map is diffible as a function of x, find its partial derivative. at a fixed x repeat for y. show these partials are continuous in some neighbourhood of a point, and conclude from thingy-bob's criterion that the function is differentiable at that point with derivative the matrix of partial derivatives.

    the second is a simple use of the chain rule.
     
  6. Jan 29, 2004 #5

    HallsofIvy

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    Tom, some textbooks use "the" derivative of a function of several variables to mean the gradient.

    Rajvirnijjar appears to be studying general derivatives of functions from Rn to Rm where that convention is widely used.

    By the way, even in "elementary" calculus of functions of several variables, showing that the partial derivatives exist is not sufficient to show that a function is "differentiable". If I remember correctly, showing that the partial derivatives are continuous is.

    As for the commas, f(u,v,w) = (eu-w, cos(v+u) + sin(u+v+w))
    is a function from R3 to R2 and
    g(x,y) = (ex, cos(y-x), e-y) is a function from R2 to R3- so it does make sense to talk about fog: a function from R2 to R2. It's derivative would be a 2 by 2 matrix (more correctly, the derivative (at a point in R2) would be a linear transformation from R2 to R2, representable by a 2 by 2 matrix in a given coordinates system.
     
  7. Jan 29, 2004 #6

    Tom Mattson

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    Ach, you're right. Demonstrating the existence is necessary and sufficient for differentiability only in the calculus of a single variable.

    Yes, you do remember correctly. I just looked it up, and that is indeed the case.
     
  8. Jan 30, 2004 #7

    HallsofIvy

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    Now that I have the time:
    You are given that f(u,v,w) = (eu-w, cos(v+u) + sin(u+v+w)) and
    g(x,y) = (ex, cos(y-x), e-y).

    fog(x,y)= f(ex, cos(y-x), e-y)
    = (exp(ex-e-y), cos(cos(y-x)+ex+ sin(ex+ cos(y-x)+ e-y)

    You could find Dfog(0,0,0) by differentiating that but that way leads to madness! I suspect your teacher's point is to show how much simpler it is to use the chain rule.

    Dg= [ex 0 ]
    [sin(y- x) -sin(y-x)]
    [0 -e-y]

    and Dg(0,0)= [1 0]
    [0 0]
    [0 -1]

    Df= [eu-w 0 -eu-w]
    [-sin(v+u)+cos(u+v+w) [-sin(v+u)+cos(u+v+w) cos(u+v+w) ]

    Caution! When x= y= 0, u= e0= 1, v= cos(0-0)= 1, and
    w= e-0= 1.

    Df(1,1,1)= [1 0 1 ]
    [-sin(2)+ cos(3) -sin(2)+ cos(3) cos(3)]

    Dfog(0,0) is the product of those matrices:
    [1 1 ]
    [-sin(2)+ cos(3) -cos(3)]
     
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