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Derivatives question

  1. Nov 8, 2007 #1
    1. The problem statement, all variables and given/known data
    hi, I am supposed to differentiate the following:

    f(2x+g(x))

    I think the problem should be solved as follows:

    f`(2x+g(x)) * (2x+g(x))` * g(x)`



    but, my ta said to do the following:

    f`(2x+g(x)) * (2x+g(x))`


    please tell me, physicsforums, which is the correct way?

    why?
     
  2. jcsd
  3. Nov 8, 2007 #2

    Avodyne

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    Your TA is right.

    Suppose you had f(h(x)). What would you say the derivative (with respect to x) is?

    Got an answer?

    OK, now let h(x)=2x+g(x).
     
  4. Nov 8, 2007 #3
    "Suppose you had f(h(x)). What would you say the derivative (with respect to x) is?"

    "OK, now let h(x)=2x+g(x)."

    I am confused, but luckily I know where I am confused so I guess there is hope.

    f`(h(x)) where h(x) = 2x+g(x)

    wouldn't we have: df/dh * dh/dg * dg/dx ???
     
  5. Nov 8, 2007 #4

    Avodyne

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    Let me try that again. Consider f(x^2). What is the derivative with respect to x?
     
  6. Nov 8, 2007 #5
    f` * (x^2)`
    ?
     
  7. Nov 9, 2007 #6

    Avodyne

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    OK, good. A clearer way to write this is f'(x^2)*(x^2)' And of course (x^2)'=2x.

    Now consider a more general function h(x) as the argument of f: f(h(x)). The derivative of this with respect to x is f'(h(x))*h'(x).

    Now, h(x) can be *any* function. It could be, for example, x^3. Or it could be x^3 + 2x. It doesn't matter what it is; we ALWAYS apply the formula f'(h(x))*h'(x).

    So, if h(x)=g(x)+2x, we STILL apply the SAME formula. We don't multiply again by g'(x).
     
  8. Nov 9, 2007 #7
    yeah, but I thought because it has another function inside of it (g(x)), we apply the chain rule again. sort of like we do with sin(cos(sinx)))
     
  9. Nov 9, 2007 #8

    Avodyne

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    No, because g(x) is not inside of g(x)+2; that would mean something like f(g(g(x))+2x), with an extra "nesting".
     
  10. Nov 9, 2007 #9
    this "nesting" is where I am confused? how exactly can I determine if there is a "nesting"?
     
  11. Nov 9, 2007 #10

    HallsofIvy

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    This is an example of "nesting". The second sin(x) is "inside" (an argument of) the function cosine (and the whole assembly is a function of sine again).
    [sin(cos(sin(x)))]' = cos(cos(sin(x)))[-sin(sin(x))][cos(x)]

    If instead, you had sin(cos(x)+ sin(x)), the second sine is NOT "nested"- it is not an argument of the function cosine. The sum of the two functions is an argument of the sine function.
    [sin(cos(x)+ sin(x))]'= cos(cos(x)+ sin(x))[-sin(x)+ cos(x)]
     
  12. Nov 9, 2007 #11
    but in : f(2x+g(x))

    since g(x) is inside the argument of f(x) isn't g(x) nesting?
     
  13. Nov 9, 2007 #12

    HallsofIvy

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    It is not just g(x) but the whole 2x+g(x) that is inside f- you need to use the chain rule once, not twice: df(2x+g(x))/dx = (df/du)(du/dx) where u= 2x+ g(x). that is df/dx=f'(2x+g(x))(2+ g'(x)), exactly the same as the f`(2x+g(x)) * (2x+g(x))` your TA said- as long as you understand the " ' " to mean differentiation with respect to "the" variable- 2x+g(x) in f' and x in (2x+g(x))'. The point is that you have already used the chain rule in muliplying by (2x+ g(x))"- there is no reason to multiply by g'(x) again.
     
  14. Nov 9, 2007 #13
    so we would only need to use the chain rule a second time if there was something "nesting" inside of g(x)?
     
  15. Nov 9, 2007 #14

    Avodyne

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    Yes, if the argument of g(x) was something other than x, then we would need the chain rule a second time.
     
  16. Nov 9, 2007 #15
    so it doesn't matter what g(x) is... it only matters what is in the argument of g(x)

    g(x) could = (x^3+(x-1)^1/2 - 35x^14)^5/3 ... etc...

    as long as only x is in the argument?
     
  17. Nov 9, 2007 #16

    Avodyne

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    Yes!
     
  18. Nov 9, 2007 #17
    Ok, so to sum things up.... if we have:

    f(2x+g(x))

    and g(x)=(x^3+(x-1)^1/2 - 35x^14)^5/3 (copied from the last hypothetical)

    then f`(2x+g(x))= f`(2x+g(x)) * (2x+g(x))` =
    f`(2x+g(x)) * 5/3(x^3 +(x-1)1/2 - 35x^14)^2/3
     
  19. Nov 10, 2007 #18

    Avodyne

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    The first line is correct. But then you did not compute (2x+g(x))' correctly.

    (2x+g(x))' = 2 + g'(x), and

    g(x) = h(x)^5/3, where h(x) = x^3+..., so

    g'(x) = 5/3 h(x)^2/3 * h'(x)
     
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