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Differeation by the quotient.

  1. Jul 10, 2009 #1
    im working on an equation using the quotient rule.

    this is the equation.


    8(5+x)/16+X^2 and i have to find the f'=

    i know 16+x^2 = 2x

    im not sure about the 8(5+x) it can not be arctan(x)

    could you please help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 10, 2009 #2

    tiny-tim

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    Hi morbello! :wink:

    (try using the X2 tag just above the Reply box :wink:)
    You mean (16+x2)' = 2x.

    And (8(5+x))' = 8(5+x)' = … ? :smile:
     
  4. Jul 10, 2009 #3
    so it is In(1+x) for 8(5+x)
     
  5. Jul 10, 2009 #4

    HallsofIvy

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    What is "In"? If you mean ln, No, the derivative of 8(5+x) is certainly NOT ln(x). Why would you think so?
     
  6. Jul 10, 2009 #5
    You are wondering about the derivative of 8(5+x)? I'm not for sure why you were thinking about arctan or ln. [8(5+x)]'=[40+8x]'=8. The quotient rule is:
    [tex]\left( \frac{f(x)}{g(x)} \right)' = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2} [/tex]

    In your case f(x)=8(5+x) and g(x)=16+x2.
     
  7. Jul 10, 2009 #6

    tiny-tim

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    uh?? :confused:

    are you thinking that (1/f(x))' = ∫f(x) ? It isn't! :rolleyes:
     
  8. Jul 10, 2009 #7
    yes i understand im looking for the f'(x) would it be as simple as x=1 and f'(x)=1
     
  9. Jul 10, 2009 #8
    I am totally confused by your questions and responses (it looks like all of us are). You are going to have to elaborate, otherwise we are just guessing as to what the problem is.
     
  10. Jul 10, 2009 #9
    it is wrote down that the f(x)=8(5+x) and i am looking for the f'(x)

    im thinking it is , as on my differentation table it has f(x)=x and f'(x)=1
     
  11. Jul 10, 2009 #10

    tiny-tim

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    That's correct! :smile:

    So (8(x + 5))' = ?​
     
  12. Jul 10, 2009 #11
    its the same i think ,thank you for helping.
     
  13. Jul 10, 2009 #12

    Cyosis

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    Don't say it's the same but give us the expression. If you're thinking that the derivative is the same as the derivative of x you're mistaken.
     
  14. Jul 10, 2009 #13
    No it's not. I did it for you in post #5. Distribute the 8 and take the derivative. Do you understand how to take simple derivatives (excluding the product and quotient rules)?

    I'm sensing some real misunderstanding on how to take derivatives, so if you can post what you are thinking, then it will allow us to help you out. You need to elaborate in your responses, because I am having a hard time understanding where you are coming from. Do you understand how to take derivatives? Do you understand what the quotient rule means? I posted it in post #5.
     
    Last edited: Jul 10, 2009
  15. Jul 10, 2009 #14
    im not sure what you mean i have jumped a'level maths and am studying this but im working around f(x)= 8(5+x) and f'(x)=1
     
  16. Jul 10, 2009 #15

    Cyosis

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    No that is wrong. Can you distribute the 8 so that the brackets disappear from your expression?
     
  17. Jul 10, 2009 #16

    tiny-tim

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    morbello, let's take this slowly …

    yes, (x)' = 1 … so what's (8x)' ?​

    (if you don't know immediately, then use the product rule :wink:)
     
  18. Jul 10, 2009 #17
    [8(5+x)]'=[40+8x]'=8

    you mean this. but im still not sure how you got the =8

    40+8x=8 could you explain.
     
  19. Jul 10, 2009 #18

    Cyosis

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    Do you know what the derivative of a constant is, in this case 40? Try to calculate (40)' and (8x)'.
     
  20. Jul 10, 2009 #19
    i think i follow you it will only be the 8x that will be used in the rest off the question. which would be 8x=8
     
  21. Jul 10, 2009 #20

    Cyosis

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    Why would only the 8x be used?
     
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