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Puzzle: (d/dt)[f(x+t)] v.s. f '(x+t)

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Are (d/dt)[f(x+t)] and f '(x+t) always equal?

    2. Relevant equations
    3. The attempt at a solution
    For (d/dt)[f(x+t)], I would interpret it as first evaluate at x+t, then differentiate with respect to t.
    And the standard interpretation of f '(x+t) is f '(x+t) = (d[f(x)]/dx) |x=x+t (first differenate, then evaluate at x+t, so the derivative itself is treated as a function).

    I've tried a few functions, and they both expressions give the same answer which is quite surprising to me...
    e.g.) f(x)=x^2, f '(x)=2x, f '(x+t)=2(x+t)
    f(x+t)=(x+t)^2, (d/dt)[f(x+t)]=2(x+t)

    e.g.) f(x)=ln(x), f '(x)=1/x, f '(x+t)=1/(x+t)
    f(x+t)=ln(x+t), (d/dt)[f(x+t)]=1/(x+t)

    However, these examples may just be special cases. It's just does not seem clear to me that these two expressions are always equal.

    Is there any way to prove rigorously that these two expressions indeed always give the same answer? (or disprove it using a counterexample?)

    Thanks for any help!
  2. jcsd
  3. Feb 5, 2009 #2


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    usually f'(x+t)=df/d(x+t) which with the chain rule means:
    But it really depends on the notation, and connotation.
  4. Feb 5, 2009 #3
    So are (d/dt)[f(x+t)] and f '(x+t) ALWAYS equal?

    I don't see how the chain rule can be applied here. The chain rule says something like dy/ds=(dy/dx)(dx/ds) which doesn't seem to apply in our situation.
  5. Feb 5, 2009 #4


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    Try making a variable substitution to make things easier; Let [tex]u\equiv x+t[/tex]. Then you want to see whether


    What must [tex]\frac{du}{dt}[/tex] be for that to occur?
  6. Feb 5, 2009 #5


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    for example:
    is this example simplifiies this matter?
  7. Feb 5, 2009 #6


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    Not a very good example for two reasons:

    (1)[tex]e^{iwt-kx}[/tex] is a function of [itex]iwt-kx[/itex] not [itex]x+t[/itex]

    (2)You are implicitly assuming that [tex]\frac{dx}{dt}=0[/tex]
  8. Feb 6, 2009 #7
    I don't get it. So is there a counterexample or is the statement true in general? (I am not familiar with complex numbers i)
  9. Feb 6, 2009 #8


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    Here's a very simple example to consider:


    [tex]\implies f'(x+t)=1[/tex]

    [tex]\implies \frac{d}{dt} f(x+t)=\frac{dx}{dt}+1=\frac{dx}{dt}+f'(x+t)[/tex]

    In this case, [tex] \frac{d}{dt} f(x+t)=f'(x+t)[/tex] is only true if [tex]\frac{dx}{dt}=0[/tex].

    You can (and should!) use the chain rule to see that the same condition applies to all functions of the form [itex]f(x+t)[/itex]
  10. Feb 6, 2009 #9
    Thanks, but there are still a few points that I don't understand:

    f '(x+t) = 1 <----here what are you differentiating with respect to?

    [tex]\implies \frac{d}{dt} f(x+t)=\frac{dx}{dt}+1[/tex] <---what principle are you using in this equality? And when we have a function f that is a function of both x and t (i.e. multivariable function), why can we still have d/dt? Shouldn't it be a partial derivative?

    If the original statement is true only when dx/dt=0, that means it's true whenever the variables x and t are independent, right?
  11. Feb 6, 2009 #10


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    This is specifically for the example f(x+t)=x+t. Surely you know that [tex]\frac{d}{dt} (x+t)=\frac{dx}{dt}+1[/tex]?

    Typically, one would treat f(x,t) as a multivariable function; and f(x+t) as a single variable function with that variable being (x+t).

    Normal derivatives are still defined for multivariable functions, so I see no reason why it should be a partial derivative (unless of course the original question actually asked if [tex]\frac{\partial}{\partial t} f(x+t)[/tex] was always equal to f'(x+t))

    Yes, but I only showed it was the case for the specific example where f(x+t)=x+t. It's your job to show that it is true for all functions of the form f(x+t).
  12. Feb 6, 2009 #11
    It might be worth imposing limits on your functions to see if they do indeed remain convergent to a particular value. By doing that you can see if this is always the case. Try proof methods.
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