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Multiple Choice-Derivatives-Please Check?

  1. Jun 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Which of the following is not the same as the derivative of y with respect to x, if y= f(x)?

    (a) lim as [itex] Δx [/itex] -> 0 of [itex] \frac{Δy}{Δx} [/itex]
    (b) lim as h--> 0 of [itex] \frac {f(x+h) -f(x)}{h} [/itex]
    (c) lim as [itex] x_1 [/itex] -> x of [itex] \frac {f(x_1)-f(x)}{(x_1-x)} [/itex]
    (d) lim as h-> 0 of [itex] \frac {f(x) - f(x-h)}{h} [/itex]
    (e) lim as [itex] Δx-> 0 [/itex] of [itex] \frac {f(Δx)}{Δx} [/itex]
    (f) lim as h-> 0 of [itex] \frac{f(x)- f(x+h)}{-h} [/itex]

    2. Relevant equations



    3. The attempt at a solution

    I think C is the correct answer. I immediately ruled out a, b, and e by recognizing them as accurate forms. I thought F was wrong at first, but then I noticed that the original form was changed (negative sign), so that is why the h is negative in the denominator. As for D, I just thought it didn't seem right at all. To be honest, I thought D and C are both incorrect, but the question seems to be asking me for one incorrect answer. Thank you.
     
  2. jcsd
  3. Jun 15, 2013 #2
    For C, I suggest you observe what happens if you let x1 = x+h.

    For D, I suggest you observe what happens if you let x1=x-h.

    I also suggest that you not immediately rule anything out, and try making substitutions and algebraic manipulations. Only one of these is incorrect, and you didn't seem to consider it.
     
    Last edited: Jun 15, 2013
  4. Jun 15, 2013 #3
    Thank you so much for your insight! I wouldn't have thought of it that way. Clearly, D is the right choice then.
     
  5. Jun 15, 2013 #4
    No, D is a correct expression for the derivative. Replace h with -h in that expression to see why (the limit is double sided so this is allowed). Take another look at E.
     
  6. Jun 15, 2013 #5
    No,

    For D, let x1=x-h.
     
  7. Jun 15, 2013 #6
    Okay, I was able to rule out C with your help via algebraic manipulation.
    As for A and B, I think that both are standard forms of the derivative, so I will put that aside for now.
    For D, I understand now how the -h makes it work.
    E: I think I should replace Δx with x + h ?
    Then it would make the expression be f(x+h)/(x+h), which I think is wrong.
    F: I still think f is correct.

    Thank you!
     
  8. Jun 15, 2013 #7
    You are right that E is not a correct form for the derivative. Why are you replacing Δx with x+h though?
     
    Last edited: Jun 15, 2013
  9. Jun 15, 2013 #8

    For E, instead of a replacement, I suggest you observe the claim it is making.

    Basically, I think you'll agree that E hinges on the idea that:

    (Delta)f(x) = f((delta)x)

    But, remember that delta x is just the difference between two x's. So let's replace delta x with (x1-x2).

    Delta f(x) = f(x1-x2)

    Or

    f(x1)-f(x2) = f(x1-x2)

    But, that is merely a property of linear functions, not any function. So E is only true for a very limited class of functions, those with a constant slope.

    Sorry for no latex, sent from phone. Hope this helps.
     
  10. Jun 15, 2013 #9
    Your explanation was so clear. Thank you so much 1MileCrash. I truly appreciate it.
     
  11. Jun 15, 2013 #10
    That was a idiotic thing I did, which I only realized after I'd written it. Delta x is the change in x, but I wrote this instead. I do understand the concept of delta x though.
     
  12. Jun 15, 2013 #11
    You're welcome, good luck. :)
     
  13. Jun 15, 2013 #12
    Thanks! :)
     
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