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Chain rule with table of values

  1. Sep 21, 2006 #1
    ok firstly i understand how the chain rule works however im given this table

    x f '(x) g(x) g'(x) h'(x)
    0 7 2 ___ 32
    2 8 0 -3 ___

    im told to fill in the blanks where h(x)=f(g(x^2-x))

    this problem has become an annoyance at first i assumed to solve i would take f ' * g '(x)*x ' however this doesnt work and no form of multiplication will give me the correct answer when i multiply them together so im guessing there is addition or subtraction involved. Please help as this is the only problem i am stuck on and the test is tom.
     
  2. jcsd
  3. Sep 21, 2006 #2

    nazzard

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    Hello trajan22,

    you didn't apply the chain rule correctly. Maybe it would help you to see where you've made the mistake, if you write down the arguments of all functions as well.

    Regards,

    nazzard
     
  4. Sep 21, 2006 #3
    i think i found my error in the application if im right then it should be
    f '(g(x))* g' (x)* (x')
    this would translate to for the second row
    8*2*-3*2
    however this doesnt bring me to the right answer i think the problem is in the x^2*x do i sub in 2 for this for every part of the operation involving x or do i take the table value ex. g'(x)=-3 or g'(x)=-3(2^2-2)or -6 im thinking this is where the problem lies. what are your thoughts
     
  5. Sep 21, 2006 #4

    nazzard

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    You still got the arguments/variables wrong. You could try to rename (x^2-x) to y(x). Then h(x) would be: h(x)=f(g(y(x))). Can you see now that you didn't apply the chain rule correctly to the y(x) part and that the arguments should read y(x) and not simply x?
     
  6. Sep 21, 2006 #5
    sorry im still havin trouble with it because since g(x) is 0 then the outcome would be 0 because you multiplyby 0 obviosly this isnt the case...

    but with the new argument is it

    8*(-3*(2))*(0(2))*-3(2)(3)*(3)??
     
    Last edited: Sep 21, 2006
  7. Sep 21, 2006 #6

    nazzard

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    That doesn't seem to be correct.

    Before plugging in any values, please make sure you have the correct answer for h'(x).

    h'(x)=f'(g(y(x))*g'(y(x))*y'(x).

    with y(x)=x^2-x.

    What is y'(x)?
     
  8. Sep 21, 2006 #7
    y'(x) should be 2 right? because x^2-x would be 2^2-1 or just 3? because y'
    is 2x-1
     
  9. Sep 21, 2006 #8

    nazzard

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    y'(x)=2x-1

    Now, as a result we have:

    h'(x)=f'(g(x^2-x))*g'(x^2-x)*(2x-1)

    What is h'(0)? To find out, you just replace every x with 0, right?

    So you'll get:

    h'(0)=f'(g(0^2-0))*g'(0^2-0)*(2*0-1)
    h'(0)=f'(g(0))*g'(0)*(-1)
     
  10. Sep 21, 2006 #9
    i follow what you are saying but still end up with a result of zero when it should be -21 im missing something obvious...ps thank you for all your help i really appreciate it
     
  11. Sep 21, 2006 #10

    nazzard

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    No problem. :smile:

    I'll try to cover it step by step.

    We have:

    h'(0)=f'(g(0))*g'(0)*(-1)

    We have a look at the table and find:

    h'(0)=32, g(0)=2

    So we put that information in our equation and get:

    32=f'(2)*g'(0)*(-1)

    luckily the table also provides us with the value for f'(2), it's 8.

    So we get:

    32=8*g'(0)*(-1).

    You can solve for g'(0) and write the value into your table.

    Can you go from there and solve the equation for h'(2)?
     
  12. Sep 21, 2006 #11
    well now i see how the answers are brought about but i didnt realize that you could deal with this by leaving the row such that f'(0)=8 but f'2 =7 what is the reasoning behind this if x is a different value because the way i thought it was i f'0=7 and f'2 =8 but this is not the case so im confused on the underlying concept....also for 32=8*g(-1) i get -4 but the answer is supposed to be just 4 32=-8g -4=g im also getting the opposite sign for h as 21 not -21
     
  13. Sep 21, 2006 #12

    nazzard

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    Hmmm, starting with this table and h(x)=f(g(x^2-x)) I don't get 21 or -21 as an answer for h'(2). :frown:

    x__f'(x)__g(x)__g'(x)__h'(x)

    0___7____2_____??___32

    2___8____0____-3____??

    Maybe I've done something wrong along the way? I'll have another look.
     
  14. Sep 21, 2006 #13
    well this is how i figured it for h'2 since f'=gx it is f'=2(0^2-0)so f' is 0 so therefore for at f'0 according to table it is 7 then 7*g' is -21(correct answer) but for some reason -21*-1 gives 21 this is the way i reasoned it hope you can follow.
     
  15. Sep 22, 2006 #14

    nazzard

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    Ok I can't seem to follow you here and I had problems with your notation in another post as well. What does f'=gx and f'=2(0^2-0) mean? :redface:

    for h'(2) I get:

    h'(2)=f'(g(2^2-2))*g'(2^2-2)*(2^2-1)
    h'(2)=f'(g(2))*g'(2)*(3)
    h'(2)=f'(0)*(-3)*(3)
    h'(2)=8*(-3)*(3) *wrong*

    :uhh:

    *edit: I get h'(2)=7*(-3)*(3) since f'(0)=7.*
     
    Last edited: Sep 22, 2006
  16. Sep 22, 2006 #15
    oh sorry about that now that i look at it again f'=g(x) doesnt make sense....but the 2is the value at g(x) and 0 is the value at x so i input this into 2*(x^2-x)which is g'(x^2-x)
    which gives the value of 0
    so that makes it f'(0) so i took this value from the table at f'(0) to be 7and multiplied it by g'(x) and returned with -21 i then multiplied by -1 as that is what x' is therefore i got 21...i did the same operation for the first colum to get g

    i hope this you can follow this ...

    WELL CANCEL THIS, IM GETTING THE RIGHT NUMBER BUT I REALIZED THAT IT IS MULTIPLIED BY 3 AT THE END BECAUSE 2X-1=3 WHEN X IS 2 SO THIS INVALIDATES MY ARGUMENT. AND WOULD CHANGE IT TO 63
     
    Last edited: Sep 22, 2006
  17. Sep 22, 2006 #16

    nazzard

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    My final table looks like this:

    x__f'(x)__g(x)__g'(x)__h'(x)

    0___7____2____-4____32

    2___8____0____-3___-63

    I'm still not sure where the mismatch with the signs comes in. :uhh:
     
    Last edited: Sep 22, 2006
  18. Sep 22, 2006 #17
    right thats what im getting and also a friend of mine who worked the problem ...my only guess at this point that the answer sheet is wrong with the h'(x) value being 21 and the g'(x) value being 4. oh well...you still helped with the comprehension of this style of problem so thanks again...
     
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