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Chain rule proof

  1. Jan 26, 2009 #1
    Hello! I got one question for you.

    How come that [tex](f \circ g)'(x) = f'(g(x)) g'(x)[/tex] ?

    Since [tex](f \circ g)'(x)=f(g(x))'[/tex] , [tex]f'(g(x))=f'(g(x)) g'(x)[/tex]. And now we can rewrite the equation like [tex]1=g'(x)[/tex]

    I don't understand that part.

    Also I don't understand why the flawed proof of the chain rule is incorrect?

    [tex]y'=\lim_{dx \rightarrow 0}\frac {dy}{dx} = \lim_{dx \rightarrow 0}\frac {dy} {du} \cdot\frac {du}{dx}=\lim_{du \rightarrow 0}\frac {dy}{du} \cdot \lim_{dx \rightarrow 0}\frac {du}{dx}=f'(u)\cdot u'(x)[/tex]

    Thanks in advance.

    Regards.
     
  2. jcsd
  3. Jan 26, 2009 #2

    CompuChip

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    Hi Dyawol.

    That's precisely what the chain rule says, I will prove it below for you.

    I don't really understand either... what are you trying to do here? Your notation is confusing you, the [tex]f'(g(x))[/tex] on the left hand side is not the same as that on the right hand side...

    Well actually it gives the correct formula (y' = f'(u) u'(x)) assuming y(x) = f(u(x)), although what is written down is nonsense.
    [tex]y' = \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}[/tex]
    is correct but I don't see what the limits are doing there, and actually it is just another way of writing f'(u) u'(x); so there is no proof here, you are just stating the chain rule.

    Perhaps it is helpful to first consider an example of how the chain rule works. Suppose you have
    [tex]f(x) = (3x^2 + 6x - 9)^2[/tex]
    and you are asked for f'(x).
    Then you note that you don't know how to do this derivative (after going through your familiar list of derivatives of elementary functions, product rule and quotient rule) but that it looks a lot like a quadratic function. If you set [itex]u = u(x) = 3x^2 + 6x - 9[/itex] then you can simply write [itex]f(x) = f(u(x)) = u(x)^2[/itex], which we usually in a slight shorthand / notational abuse write as [itex]f(u(x)) = u^2[/itex] or [itex]f(x) = u^2[/itex] (which is slightly confusing perhaps, because it is not clear that there is still an x involved). Now this we know how to differentiate: the derivative of [itex]u^2[/itex] is just 2 u. So we would write
    [tex]\frac{df}{du} = 2u [/tex]
    to indicate that if u were the variable we were interested in, the derivative of f would be 2u. But we don't want df/du, we want df/dx. The chain rule tells us, that what we wanted to calculate, df/dx, is given by
    [tex]\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx},[/tex]
    i.e. we still have to multiply 2u by the derivative of u with x as the variable. Recalling that u was [tex]3x^2 + 6x - 9[/tex], we can apply our standard repertoire of derivation tricks and get
    [tex]\frac{du}{dx} = 6 x + 6[/tex]

    So, putting it all together, the answer we wanted it
    [tex]f'(x) = \frac{df}{dx} = \frac{df}{du} \frac{du}{dx} = (2u) \cdot (6x + 6)[/tex]
    where we now have to write u back in terms of x:
    [tex]f'(x) = (3x^2 + 6x - 9) \cdot (6x + 6)[/tex]
    which you could simplify to
    [tex]f'(x) = 18(x^2 + 2x - 3)(x + 6).[/tex]

    Do you understand now the derivations with respect to x and u, and the notation
    [tex]\frac{df}{du} \text{ and } \frac{df}{dx}?[/tex]
    Then you have to get used to the "mathematical" shorthand, where we usually write f'(x) if we mean df/du, u'(x) for du/dx; we can make up notations like f'(u) for df/du but I urge you to use the d.../d... notation, because f'(u(x)) is very ambiguous (this is what was confusing you in the first post: does the prime in f'(u(x)) indicate derivation with respect to u or x? That is, do you mean df/dx or df/du here?)
     
  4. Jan 26, 2009 #3

    Fredrik

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    I disagree with this. f'(u(x)) can only mean df/du, since the prime is on f. If you differentiate with respect to x, the function you're taking the derivative of isn't f. It's [itex]f\circ u[/itex], so you must write [itex](f\circ u)'(x)[/itex] (or [itex]\frac{d}{dx}f(u(x))[/itex] )
     
    Last edited: Jan 26, 2009
  5. Jan 26, 2009 #4
    Thank you very much for the help CompuChip. It really helped me understand what it mean.

    But I didn't understand one thing (the notation one), since f o g (x) = f(g(x))
    So (f o g) ' (x) would probably mean f(g(x))'. Am I right?

    < where we usually write f'(x) if we mean df/du


    I think that you thought about f'(x)=df/dx

    Also I think that you missed to multiply by 2, since there is 2u. So it would be [tex]f'(x) = 36(x^2 + 2x - 3)(x + 6)[/tex]

    Thanks again for the help.

    Regards.
     
  6. Jan 26, 2009 #5
    Why do you have x + 6, if anything it should be x + 1 if you are taking the 6 out.
     
  7. Jan 29, 2009 #6

    Fredrik

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    You're right, assuming that what you mean by f(g(x))' is the value at x of the derivative of the function that takes any number y in the domain of g to f(g(y)). But you should never write f(g(x))'. The prime symbol should only appear on a function, and f(g(x)) is not a function. It's a number. The function that you have in mind can be written as [itex]x\mapsto f(g(x))[/itex] or [itex]f\circ g[/itex].

    The derivative of that function can be written as [itex](f\circ g)'(x)[/itex] or [itex]\frac{d}{dx}f(g(x))[/itex]. Note that the dx in the denominator tells us that the function we're taking the derivative of is [itex]f\circ g[/itex]. If we had been interested in the derivative of f at the point g(x), we would have written it as [itex]f'(g(x))[/itex] or [itex]\frac{d}{dg}f(g(x))[/itex].
     
    Last edited: Jan 29, 2009
  8. Jan 29, 2009 #7

    Vid

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  9. Feb 3, 2009 #8

    Fredrik

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    See also #4 and the beginning of #3 in this thread for more about differentials.
     
  10. Feb 3, 2009 #9
    I don't see what's the problem here:
    f(g(x))'=df/dx=(df/dg)(dg/dx)

    The way you prove it is by looking at [f(g(x+h))-f(g(x))/(g(x+h)-g(x))][(g(x+h)-g(x))/h], where h->0.
     
  11. Feb 3, 2009 #10

    Fredrik

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    I'm not sure if you're talking to me or the OP, but the issue in this thread isn't just how to prove it, but to understand why you can't prove it just by canceling differentials in the expression

    [tex]\frac{df}{dx}=\frac{df}{dg}\frac{dg}{dx}[/tex]
     
  12. Feb 8, 2009 #11
    You can't just cancel differentials because dg may in fact be 0, in which case the above is nonsensical. Thus the use of differentials is only a heuristic, not a proof.
     
  13. Aug 10, 2011 #12
    [tex]\lim_{\delta x \rightarrow dx}\frac{\delta y}{\delta u}\frac{\delta u}{\delta x}=\lim_{\delta x \rightarrow dx}\frac{\delta y}{\delta x}=\frac{dy}{dx}[/tex]
     
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