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Homework Help: Find the derivative

  1. Mar 23, 2006 #1
    Find f'(x) if it is known that
    [tex]\frac{d}{dx}[f(2x)] = x^2[/tex]

    I let u(x) = 2x, then
    [tex]\frac{d}{dx}[f(u)] = \frac{dy}{du} \frac{du}{dx}[/tex]
    [tex]\frac{d}{dx}[f(2x)] = 2 \frac{dy}{du}[/tex]
    [tex]\frac{dy}{du} = \frac{1}{2}x^2[/tex]
    [tex]f'(x) = \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}[/tex]
    [tex] = (\frac{1}{2}x^2) (2)[/tex]
    [tex] = x^2[/tex]
    why doesn't this work??
  2. jcsd
  3. Mar 23, 2006 #2
    u have f'(2x) = x^2

    first find f(2x) and try to write it as a function of (2x)

    and then after u do that substitute x for 2x
    you therefore find f(x) finally derive it and you obtain f'(x)

    does it make any sense?
  4. Mar 23, 2006 #3
    yeah, i think so.
  5. Mar 23, 2006 #4
    let me know what your answer is to see if you did it right
  6. Mar 23, 2006 #5
    Well, i kinda cheated and looked at the solutions manual:tongue2:
    Using the chain rule:
    [tex]\frac{d}{dx}[f(2x)] = 2 f'(2x) = x^2[/tex]
    [tex]f'(2x) = \frac{1}{2}x^2[/tex]
    then let u(x) = 2x
    [tex]f'(u) = \frac{1}{2} (\frac{u}{2})^2[/tex]
    [tex]f'(u) = \frac{1}{8} u^2[/tex]
    then substitute x for u:
    [tex]f'(x) = \frac{1}{8} x^2[/tex]

    Is there an easy to understand explanation why my original method did not work?
  7. Mar 23, 2006 #6
    hehe good

    my method was simple: f'(2x) = x^2 --> f(2x) = x^3 / 3

    we multiply the denominator and numerator by 8 --> f(2x) = 8x^3 / 24

    ---> f(2x) = (2x)^3 / 24 ---> f(x) = x^3 / 24 --> f'(x) = x^2 / 8
  8. Mar 23, 2006 #7
    your mistake was taking partial derivatives

    u write f'(x) = dy/dx = (dy/du)/(du/dx)

    since y and u are both functions of x u can not apply the partial derivative (the chain rule formula)
  9. Mar 23, 2006 #8
    Thanks A_I_, I like your method :tongue:

    can you expand on what you said my mistake was? I'm not sure I understand.
  10. Mar 23, 2006 #9
    the chain rule is used when u have:

    f(x,y) and x=g(t) and y=r(t)

    but since your f and your u are both functions of x,
    thus you can not use the chain rule and you can not say:

    f'(x) = (dy/du)*(du/dx)

  11. Mar 23, 2006 #10
    Yeah, i think I understand.

    Can you explain that function syntax? I've seen it before, but I've always learned the simple f(x) not f(x,y)

    and for composite functions (using the chain rule), f(x) = g(r(x))
  12. Mar 23, 2006 #11
    what do you exactly want to know?
    do you want an example of the chain rule?

    what you wrote: f(x)= g(r(x)) is (gor)(x) and is different from the chain rule.

    do you have an im?
  13. Mar 23, 2006 #12
    i know what the chain rule is.
    i just want to know what f(x,y) means and what "f(x,y) and x=g(t) and y=r(t)" means.
  14. Mar 23, 2006 #13
    here you go

    Attached Files:

    • bikh.JPG
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  15. Mar 23, 2006 #14
  16. Mar 25, 2006 #15


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    What do you mean by partial derivative here??? You only encounter partial derivative in multi-variable function, not 1 variable function like f(x)!!!!
    NOOO!!!!!! What do you mean by this??? I am TOTALLY lost!!!
    Why can't you use the chain rule in that case? Am I misssing something? :confused:
    Please look back at your cal textbook, see the part that covers the chain rule.
    [tex]\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}[/tex]
    [tex]\frac{df(2x)}{dx} = \frac{df(2x)}{d(2x)} \times \frac{d(2x)}{dx} = 2 f'(2x)[/tex].
    Now, let's do your problem in a slightly different way:
    [tex]\frac{df(2x)}{dx} = 2 f'(2x) = x ^ 2[/tex]
    [tex]\Rightarrow f'(2x) = \frac{x ^ 2}{2}[/tex]
    Now, let y = 2x, we have:
    [tex]f'(y) = \frac{(2x) ^ 2}{8} = \frac{y ^ 2}{8}[/tex]
    Now, what's f'(x)?
    Can you go from here? Is there anything unclear? :)
    Whoops, looking back at some previous posts of this thread, I saw that you've looked through the manual.
    The only error you made is that you were trying to find dy / dx, which means you were finding d(f(2x)) / dx, not d(f(x)) / dx (which means the same as f'(x)).
    Is there anything unclear, endeavor? :)
    Last edited: Mar 25, 2006
  17. Mar 25, 2006 #16
    for the first part, thats what i was trying to tell him,
    that we only use the chain rule when we have a multivariable function, which is not the case.
    As for the second part, i am not sure about it, because i know you can only use the chain rule with multivariable function.
    Did u consider x and 2x to be two different variables, if yes, then it works.
  18. Mar 25, 2006 #17


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    What do you mean???
    Chain rule can be used for both multivariable functions, and 1 variable function.
    You must have studied [tex]\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}[/tex] BEFORE studying multi-variable functions, no?
    As I told you before, you should re-read your calculus textbook, just look up the chapter for Chain rule (or you can just click on the link) for one variable function. That won't do you any harm, I promise. :wink:
    Last edited: Mar 25, 2006
  19. Mar 25, 2006 #18
    I get what you've said.
    Thanks for the explanation
  20. Mar 25, 2006 #19
    So when in my first post I wrote:
    that really means
    [tex]\frac{d f(2x)}{du}[/tex]
    since y = f(2x) ?

    and then later on, when I wrote:
    [tex]\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}[/tex]
    I was actually finding f'(2x), which was already given??
  21. Mar 25, 2006 #20


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    Yes, this is correct. :)

    Nah, this is not correct, you are finding:
    [tex]\frac{df(2x)}{dx}[/tex], not f'(2x).
    [tex]\frac{df(2x)}{dx} \neq f'(2x)[/tex].
    To find f'(2x), you must find: [tex]\frac{df(2x)}{d(2x)}[/tex].
    Can you get this? :)
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