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F primes derivative

  1. Apr 17, 2014 #1

    gingermom

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    1. The problem statement, all variables and given/known data

    d/dx[f(2x)]=f'(x) and f'(1)=1 find f'(2)

    2. Relevant equations



    3. The attempt at a solution
    so I think this means the derivative of "f(2x)" equals f'(x)
    if I find the derivative using the chain rule I would get
    f'(2x)2= f'(x)
    so f'(2x) = f'(x)/2



    Here I am at a loss as to how to use the information about f'(1)=1 to find f'(2).

    Do I just substitute 1 for f'(x) and then 1 for x in the f'(2x)? That would make the answer 1/2.

    As you would have f'(2)=1/2 Is that correct? Am I trying to make this problem harder than it is?
    Is the value of f'(2) the only value I could determine from this data? i.e. I could not determine the value of f'(3).
     
    Last edited by a moderator: Apr 17, 2014
  2. jcsd
  3. Apr 17, 2014 #2

    BiGyElLoWhAt

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    Rewrite your first (given) equation in terms of all f primes. This is actually more of a logic problem than a math problem in my opinion.

    Hint* Let x=1

    I think you can take it from here.
     
  4. Apr 17, 2014 #3

    gingermom

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    Yes - a logic question is probably a more appropriate term. I am afraid I am missing something. I thought what you suggested is what I did to get the answer. If that is not the case, where are you suggesting I substitute 1 for x? I feel like I am missing a reason why or that I made a leap to a connection that is not quite right.

    So, are you suggesting my answer of f'(2) =1/2 is incorrect? Or that my logic is incorrect, or both?

    A little more direction would be appreciated - just a first year Calculus student trying to get a grip on this.
     
  5. Apr 17, 2014 #4

    BiGyElLoWhAt

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    well if f'(x)=f'(2x), and f'(1)=1 what does f'(2)=? I know I'm just restating the question here. Substitute 1 for x in the first eqation (f'(x)=f'(2x)). Post back what you get when you sub it in if you haven't figured it out yet.
     
  6. Apr 17, 2014 #5

    gingermom

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    Doesn't that second set of brackets change what we are doing. I thought the d/dx f[f(2x) would mean f'(f(2x)) not f'(2x), so wouldn't that be f'(x) = f'[f(2x)] , not f'(x)=f'(2x). Which is why I applied the chain rule. Maybe I just don't understand the nomenclature.
     
  7. Apr 17, 2014 #6

    BiGyElLoWhAt

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    I think you're misunderstanding what's going on here:

    ##\frac{d}{dx}f(x) =f'(2x)##

    Your book or instructor is just using different notations (probably to make sure you understand what you're doing)

    ##\frac{d}{dx}f(x)## is the same as saying ##f'(x)##

    so what you have here is this:

    ##f'(x)=f'(2x)## and you know that ##f'(1)=1## since f is a function of x, you're plugging in 1 for x. That's why I told you to sub in 1. Look what happens when you sub in 1 for x in the first equation:

    ##f'(x)=f'(2x) → f'(1)=f'(2(1))## see it yet?
     
  8. Apr 17, 2014 #7

    gingermom

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    Maybe I need a break - that just seems like f'(1) =f'(2) which would make the answer 1 ( would it not?) - but my options are 1/4; 1/2; 3/4;3/2; and 5/2.
    So my assumption was the lesson was the difference between f'(2x) and f'[f(2x)] and f'[f(2x)] meant I had to apply the chain rule d/dx2x*f'(2x) which would equal 2f'(2x). That would give me the f'(2x) which when I substituted in the 1 would give me f'(2).
     
  9. Apr 17, 2014 #8

    BiGyElLoWhAt

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    That's what I was thinking it was. Maybe I"m misunderstanding the question. Can you post a snap shot?
     
  10. Apr 17, 2014 #9

    BiGyElLoWhAt

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    Is this supposed to be:
    [itex]\frac{d}{dx}f'(2x)=f'(x)[/itex]
    ?
     
  11. Apr 17, 2014 #10

    gingermom

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    Screen shot

    I have attached a screen shot. i could not figure out how to insert it in the message. Hope that works.
     

    Attached Files:

  12. Apr 17, 2014 #11

    Mark44

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    d/dx[f(2x)] = f'(2x) * 2, using the chain rule.

    So you have 2 * f'(2x) = f'(x)
    Evaluate the equation above when x = 1. The answer is one of the options shown in the screenshot.
     
  13. Apr 17, 2014 #12

    Mark44

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    No, it's not a logic problem. It's definitely a calculus problem.
     
  14. Apr 17, 2014 #13

    gingermom

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    So was my initial analysis correct?

    f'[f(2x)] meant I had to apply the chain rule d/dx2x*f'(2x) which would equal 2f'(2x).

    f'(2x)2= f'(x)
    so f'(2x) = f'(x)/2 I then substitute 1 for x and f'(2)= 1/2?

    Thanks for stepping in - I was getting concerned that I was really missing something.
     
  15. Apr 17, 2014 #14

    Mark44

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    Mostly.
    Yes. My reason for writing "Mostly" above is to correct what you wrote as f'[f(2x)].

    The problem stated that d/dx[f(2x)] = f'(x), not f'(f(2x)).

    d/dx(<something>) means "take the derivative with respect to x of <something>.
    f'(<something>) means to evaluate the derivative f' at <something>.
    These are very different, as the first means to find the derivative, and the second means to evaluate a derivative that we have already found.

    For example, d/dx(x2) = 2x, but f'(x2) is meaningless unless somebody tells me what f is, from which I can figure out what f' is.


    Yes, that's it.
    And here you are using the notation correctly. f'(2x) means to evaluate f' (which is not known) at a number 2x.
     
  16. Apr 17, 2014 #15

    Mark44

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    I disagree. In the first post, it says "d/dx[f(2x)]=f'(x)", which, in words, is the derivative (w.r.t. x) of f(2x).

    Another member posted in this thread, and due to misconstruing what the problem was, launched off in an unproductive direction. gingermom had the correct answer in the first post.
     
  17. Apr 17, 2014 #16

    Mark44

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    OK, good, Sammy, we're both on the same page.
     
  18. Apr 17, 2014 #17

    SammyS

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    That's always good to know !!! :smile:
     
  19. Apr 17, 2014 #18

    BiGyElLoWhAt

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    Yea, I pretty much misinterpreted that whole problem. Thanks for pointing it out guys.
     
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