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Chain Rule

  1. Apr 21, 2008 #1

    These are not h/w problems but something from the class notes which I am not able to fully understand. I have two questions stated below

    We are doing chain rule and function of several variables. To explain the prof has first explained about single variables and then gone to composite functions and vector functions. But since I cannot understand single variables, no point in going to other topics

    Anyway the class note is as follows:

    The differentiability of f at xo is equivalent to
    ( f(xo+h) - (f[tex]^{'}[/tex](x0)h) ) / h -> 0 i.e. the function f is approximated to first order by function x-> f(xo) + f[tex]^{'}[/tex](x0)(x-xo) (where x = xo + h)

    The function h-> f(xo) + f[tex]^{'}[/tex](x0)h is a function of the form: a constant function (f(xo) + a linear map (f[tex]^{'}[/tex](x0)h). Such a function is called an affine function. Thus differentiability can be formulated as the statement: there is a linear map [tex]\phi[/tex]: R->R such that the affine function f(xo) + [tex]\phi[/tex](x-xo) is a first-order approximation to f at xo.

    My problem is as follows
    - it is the last line which states that differentiability can be formulated by the linear map. Can someone explain what the notes are trying to say. I have been on this for about 3 days and not able to grasp it

    Question 2:
    Not entirely related to this, but assuming you have a function f(x1,x2,x3) = x1x2x3 + x1*x3. How do I prove that it is continuously differentiable. If I prove that partial derivatives exist, will it prove that they are continuously differentiable.


  2. jcsd
  3. Apr 22, 2008 #2

    Gib Z

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    Homework Helper

    Q1. It means if a function f(x) is differentiable at a point [itex]x_0[/itex], the linear function that best approximates f(x) at x_0 is [tex]g(x) = f(x_0) + f' (x_0) (x - x_0)[/tex] ie its tangent.

    To say the differentiability of f at x_0 is equivalent to it being approximated by that tangent is somewhat of a circular argument however, as the function must be differentiable at x_0 to obtain f'(x_0) .

    Q2) What do you mean by "continuously differentiable"? If you mean repeatedly or infinitely differentiable, then showing the first partial derivatives exist only shows that they are once differentiable. You must show all further partials also exist.

    EDIT: Ahh from Hall's reply, it appears the question was referring to whether or not the partials are continuous functions in themselves. My bad for misinterpreting.
    Last edited: Apr 22, 2008
  4. Apr 22, 2008 #3


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    Staff Emeritus
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    It just says that the function f can be approximated, close to x=a, by its tangent line at x= a. At x= 1, for example, if f(x)= x2, the f '(x)= 2x so f(1)= 1 and f '(1)= 2. The tangent line to the graph of y= x2, at x= 1, is y= 2(x-1)+ 1= 2x- 1. For values of x close to 1, y= 2x- 1 will give values close to x2: better than any other linear function.

    Question 2: You are right- the fact that the partial derivatives exist does not mean they are continuously differentiable or that the original function is "differentiable". However, in this example, Your function is a polynomial. If you were to think of it as a function of one of the variables, say f(x)= abx+ bx, its derivative would be ab+b, which, being a constant, is continuous. That tells you that the original function has continuous derivatives.
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