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A derivative problem

  1. Oct 16, 2015 #1
    1. The problem statement, all variables and given/known data
    ##F'(x)##=##[2f(x)+g(x)]'##
    ##F'(0)=##
    Given g'(0)=2 and f'(0)=5


    2. Relevant equations
    image003.gif
    3. The attempt at a solution
    I can solve this if the questions is as follows
    ##F'(x)##=##[f(x)+g(x)]'##
    by applying sum rule
    but I don't know what to do about 2?##F'(x)##=##[2f(x)+g(x)]'##
    I tried to solve this as follows
    ##F'(x)##=##[2f(x)+g(x)]'##
    =##[2.f'(x)+f(x).f'(2)]## +##g'(x)]## (by applying multiplication rule) here f prime denotes derivative of
    =##[2.f'(x)+f(x)##+##g'(x)##]
    But it is wrong . Actually I do know that 2 will be taken out (it has been given in my book as solution of this problem),But I want to know why?Which rule governs this action of taking 2 out?
     
    Last edited: Oct 16, 2015
  2. jcsd
  3. Oct 16, 2015 #2

    Mark44

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    You're making this much more difficult than it actually is.
    If f'(x) = g'(x), then f(x) = g(x) + C
    IOW, if the derivatives of two functions are equal, then the two functions differ only by a constant.
     
  4. Oct 16, 2015 #3

    epenguin

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    What is the problem? None is stated.
     
    Last edited: Oct 16, 2015
  5. Oct 16, 2015 #4
     
  6. Oct 16, 2015 #5

    epenguin

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    Taking 2 out of what?

    You can't just write a formula and that is a question. It has to have some form like given ... prove that.... where this, where that... Find the other...or something.
    Best to quote the passage where this cones from, else you know what the problem is but we don't.
     
    Last edited: Oct 16, 2015
  7. Oct 16, 2015 #6
    Sorry.I have edited the OP.
     
  8. Oct 16, 2015 #7

    haruspex

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    Your error is here:
    (2f(x))' = f'(2)f(x)+2f'(x)
    It should be
    (2f(x))' = 2'f(x)+2f'(x)=(0)f(x)+2f'(x)=2f'(x)
     
  9. Oct 16, 2015 #8

    Mister T

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    The only relevant multiplication rule is that ##[2f(x)]'=2f'(x)##.

    So, ##[2f(x)+g(x)]'=2f'(x)+g'(x)##.

    The product rule tells us that

    ##[2f(x)]'=2f'(x)+(2)'f(x)=2f'(x)+0 \cdot f(x)=2f'(x)+0=2f'(x)##.

    Note that ##f'(2) \neq [f(2)]' \neq (2)'##.
     
  10. Oct 16, 2015 #9

    epenguin

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    So I guess the question is: F'(x) = ?

    I think I see what your problem is - quite a common one still with students more advanced than you.

    It looks like your have this addition formula for [ f(x) + g(x) ]' and you are regarding f and g as something rigidly fixed for all time. Or once chosen, then fixed. They aren't - nor is x for that matter. If the question had been find [ 2φ(x) + ψ(x) ]' you could probably have done it, if we want to do it very laboriously we could do it for that and then say "Let φ be f and ψ be g,..."

    Students get that x, for instance can represent anything, they do not always get the way in use in algebra etc. a moment later it can represent anything else. So I can represent just any function by f. But I can represent just any function if I want by 2f. (I might have some reason for doing that, maybe I know that later in some calculation I'm doing that later on its going to get halved and it might give me a nicer expression, or maybe it has come out of some preceding calculation. But I'm free to do it for whatever reason or none.)
     
    Last edited: Oct 16, 2015
  11. Oct 16, 2015 #10

    SteamKing

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    There are rules which govern taking the derivative when functions are multiplied by constants, when functions are added or subtracted, multiplied or divided.

    This article here shows what to do:

    https://www.mathsisfun.com/calculus/derivatives-rules.html

    Scroll down to the Table which reads 'Rules'.
     
  12. Oct 16, 2015 #11

    Mark44

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    Even in your edited post, it's still not clear what the question is. When you post a problem, write the problem statement clearly, so that there is no uncertainty in what it is you're trying to do.

    Apparently, what you're supposed to do is find F'(0), given ##F'(x)##=##[2f(x)+g(x)]'##, and g'(0) = 2, and f'(0) = 5. Using the basic rules of differentiation in the link that SteamKing gave, this is a straightforward problem.
     
  13. Oct 16, 2015 #12

    epenguin

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    Or maybe you did have a little difficulty with derivation, but the question is about functions of 0. This is just substitution.
     
  14. Oct 17, 2015 #13
    ##f'(x)##=##[f(x)]'##=##(x)'##

    here ##(x)'## means derivative of x.
    but ##f'(2)##≠##[f(2)]'##≠##(2)'##

    Right?
     
  15. Oct 17, 2015 #14

    Mark44

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    f'(x) and [f(x)]' are two ways of writing the same thing. Neither is equal to x' in general.
    We are 14 posts into what is a very straightforward problem. From other posts of yours that I have seen, you have a tendency relatively simple questions into threads that go on for 50 or more posts. Let's try not to let that happen here.

    Your original equation:
    ##F'(x) =[2f(x)+g(x)]'##

    The right side means the derivative, with respect to x, of the quantity in the brackets. Using some of the first rules your learn when differentiation is presented in calculus, how can you rewrite [2f(x)+g(x)]' ?
     
  16. Oct 17, 2015 #15
    I have understood my OP.Here OP is original problem.But I am struggling with what
    mister T wrote.Last line of #8
     
  17. Oct 17, 2015 #16

    haruspex

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    No. The derivative of f is the derivaitive of f. What makes you think it can morph into a derivative of x?
    Maybe you don't understand the f(x) notation. It means the value returned by the function f when given the input x. The derivative of f, f', is another function. f'(x) means the value of that function when given the input (or 'evaluated at') x.
     
  18. Oct 17, 2015 #17
    Should not it be f'(x) means the derivative of that function when given the input (or 'evaluated at') x.
     
  19. Oct 17, 2015 #18

    haruspex

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    f(2) is the value of the function f given the input 2. That is simply a constant number, so does not change as x changes.
    Part of your confusion comes from the fact that the prime notation, ', is a shorthand for a derivative with respect to some variable. You have to lnow from context which variable. In f'(x), it is obvious the variable is x. But in f'(2) it is no longer obvious. If it still means d/dx then f'(2) must be zero, since f(2) does not change when x changes.
     
  20. Oct 17, 2015 #19
    Then why not
    ##f'(2)##=##[f(2)]'##
     
  21. Oct 17, 2015 #20

    ehild

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    We do the operation in the brackets first. f '(2) is the value of the derivative at x=2. [f(2)] ' is the derivative of the number f(2), which is zero.

    For example, let be f(x)=x^3. Its derivative is f '(x) = 3x^2.
    The value of the derivative at x=2 is f '(2) = 3(2)^2=12.
    On the other hand, [f(2)] ' = (2^3)' =0 as 2^3 is a constant.
     
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