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Chain Rule Variation

  1. Jun 14, 2013 #1
    I have a composite function f(g(x,y)).

    When is it true that ∂f/∂g = (∂f/∂x)(∂x/∂g) + (∂f/∂y)(∂y/∂g)?

    Does g have to be invertible with respect to x and y for this to be true?
     
  2. jcsd
  3. Jun 14, 2013 #2

    Mark44

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    What you have on the right side doesn't make sense to me. You are treating g as if it were a variable rather than a function.

    ∂x/∂g is the partial of x with respect to g.
    Let's get some variables in here by assuming that z = g(x, y), and that x = h(t), y = k(t).

    So f is a function of t alone, so it makes sense to talk about df/dt.

    df/dt = ∂g/∂x * dx/dt + ∂g/∂y * dy/dt

    df/dt exists and is defined provided that all of the other derivatives exist and are defined. IOW, provided that ∂g/∂x and ∂g/∂y exist and are defined, and that h and k are differentiable functions of t.

    Note that dx/dt = h'(t) and dy/dt = k'(t).
     
  4. Jun 14, 2013 #3
    Hey Mark, thanks for the reply.

    It seems that what you are describing with the f(t) example is simply a normal chain rule. I already know that for f(x(t),y(t)) it is true that ∂f/∂t = (∂f/∂x)(∂x/∂t) + (∂f/∂y)(∂y/∂t).

    However, I am more interested in the "other way around". In the one-variable case, I mean to ask whether it is okay to say that for f(g(x)), ∂f/∂g = (∂f/∂x)(∂x/∂g).

    I'm afraid I don't understand your objection about differentiating with respect to a function. I know that taking ∂f/∂g is okay. Are you saying that ∂x/∂g is undefined? I believe I have seen similar usages in many physics problems, where ∂x/∂g is simply taken to be 1/(∂g/∂x). In this case, the change in x given a change in g, holding all other dependences of both x and g fixed, seems to be well defined. For example, if g=2x, then it's true that ∂g/∂x=2. Similarly, x=g/2 and ∂x/∂g= 1/2.

    EDIT: I just realized that the equation I wrote for the one-variable case is valid if and only if g(x) can be inverted to obtain an equation for x(g). But I'm still confused about the validity of the original problem I posed, where g(x,y) is a function of two variables. My question remains the same. Does the multivariable "inverse chain rule" hold?
     
    Last edited: Jun 14, 2013
  5. Jun 14, 2013 #4

    Fredrik

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    ∂f/∂g is not defined...unless you just mean f'.

    You have probably seen notations like
    $$\frac{d}{dx}f(g(x),h(x)) =\frac{\partial f}{\partial g}\frac{\partial g}{\partial x} +\frac{\partial f}{\partial h}\frac{\partial h}{\partial x},$$ but here ##\partial f/\partial g## just means ##D_1 f##, i.e. the first partial derivative of f. It's written with a g in the denominator because we are evaluating the function that's the result of the partial derivative operation at (g(x),h(x)), and that makes it possible to think of g(x) as the "first variable".

    I think it's a very misleading notation, since the operation of "taking the first partial derivative of f" is something that doesn't involve g in any way.
     
    Last edited: Jun 14, 2013
  6. Jun 14, 2013 #5
    Can you explain that, Fredrik? Why is ∂f/∂g not defined?

    If f=g^2, where g=3x, is it not the case that ∂f/∂g=2g=6x?

    EDIT: I'm confused, because in the normal chain rule, where ∂f/∂x=(∂f/∂g)(∂g/∂x), we see the term ∂f/∂g. Is it well-defined in this usage?
     
    Last edited: Jun 14, 2013
  7. Jun 14, 2013 #6

    Fredrik

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    I explained a bit more in an edit that that I was still typing when you posted that.

    If g is defined by ##g(x)=3x## for all x, and f is defined by ##f(x)=x^2## for all x. Then the chain rule tells us that ##(f\circ g)'(x)=f'(g(x))g'(x)=2g(x)\cdot 3 =18x## for all x.

    This specific problem can also be done without using the chain rule.

    ##f(x)=g(x)^2=(3x)^2=9x^2\Rightarrow f'(x)=18x.##

    The chain rule ##(f\circ g)'(x)=f'(g(x))g'(x)## is often written as
    $$\frac{df}{dx}=\frac{df}{dg}\frac{dg}{dx}.$$ This is a notation that I find very misleading, for the reasons mentioned in my edit of my previous post. The notation df/dg makes it look like g is somehow involved in the process of taking the derivative of f.
     
  8. Jun 14, 2013 #7

    Mark44

    Staff: Mentor

    Let's lose the ∂ notation, since the functions are single-variable here.

    df/dg = 2g
    dg/dx = 3x

    df/dx = df/dg * dg/dx = 2g * 3x = 6x * 3x = 18x2
     
  9. Jun 14, 2013 #8

    Fredrik

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    The notation f(g(x,y)) strongly suggests that ##f:\mathbb R\to\mathbb R##. So if I see the notation df/dg, or worse, ∂f/∂g, I can only interpret it as f'.

    But I couldn't even do that when I read your post, because you wrote things like ∂x/∂g on the right-hand side. This seems to rule out that we're talking about f', since a computation of f' doesn't involve any functions other than f.
     
  10. Jun 14, 2013 #9
    Ok, thanks a lot guys, your explanations really helped. I see now that since ∂f/∂g is only used as a tool in evaluating the chain rule for a variable, it has no significance on its own. Thus, it really makes no sense to try and find an expression for it! Once again, thanks!
     
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