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Differentiation Question

  1. Apr 16, 2005 #1

    aek

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    I'm having difficulties in differentiating this question y = x . (x-1)^1/2 , i keep on getting answers different to the back of the book. If someone can help, thnx
     
  2. jcsd
  3. Apr 16, 2005 #2

    dextercioby

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    Si it's

    [tex] y(x)=x\sqrt{x-1} [/tex]

    ...?Okay.What do you know about Leibniz (product) rule and the derivative of [itex] x^{n} [/itex] ?


    Daniel.
     
  4. Apr 16, 2005 #3
    Just apply the chain rule.Any of the available books can explain it to you.Maybe you are not reading and practising much.
     
  5. Apr 16, 2005 #4

    aek

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    yes im applying the chain rule, im getting an answer but the same as the real one
     
  6. Apr 16, 2005 #5

    Hurkyl

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    If you show your work, we can better help you see what you did wrong.
     
  7. Apr 17, 2005 #6

    aek

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    No the question is -
    [tex] y(x)=x\sqrt{x+1} [/tex]

    The answer is x(x+1)^1/2 +2(x+1) ALL OVER 2(x+1)

    i get an answer which is the same but with an extra x(x+1)^1/2 on the numerator. can someone help please, or does anyone get the same answer
     
  8. Apr 17, 2005 #7

    jtbell

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    To echo Hurkyl, show us how you got your answer...
     
  9. Apr 17, 2005 #8
    the way that i do it... if you dont want to use product rule
    multiply the x back into the arguement of the square root
    u get y = (x^3-x^2)^0.5
    then y' = 0.5((x^3-x^2)^(-0.5))(3x^2-2x)
     
  10. Apr 17, 2005 #9
    [tex] x\sqrt{x+1}' = \sqrt{x+1}+\frac{x}{2\sqrt{x+1}} [/tex]

    [tex] f(x)=x, g(x) = \sqrt{x+1} [/tex]

    [tex] [f(x)g(x)]' = f'(x)g(x) + f(x)g'(x) [/tex]
     
    Last edited: Apr 18, 2005
  11. Apr 17, 2005 #10

    dextercioby

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    U mean

    [tex] \left[f(x)g(x)\right]'\equiv\frac{d}{dx}\left[f(x)g(x)\right]=f'(x)g(x)+f(x)g'(x) [/tex]

    ,right?


    Daniel.
     
  12. Apr 17, 2005 #11
    I put the apostrophe on the outside meaning the derivative of both f(x)and g(x) but parantheses would be better I suppose.
     
  13. Apr 17, 2005 #12

    dextercioby

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    Mathematics is no gound for sloppy notations.Neither is physics...


    Daniel.
     
  14. Apr 17, 2005 #13

    aek

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    can someone tell me if anyone got the same answer as the book. I DID USING THE PRODUCT RULE BUDDY.
     
  15. Apr 17, 2005 #14

    dextercioby

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    Either u didn't write the book's answer correctly,or it is incorrect.

    [tex] (x\sqrt{x+1})'=\frac{2(x+1)^{3/2}+x\sqrt{x+1}}{2(x+1)} [/tex]


    Daniel.
     
  16. Apr 17, 2005 #15

    aek

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    your wrong buddy.
     
  17. Apr 17, 2005 #16

    Hurkyl

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    Can the attitude.

    And I notice you still haven't posted your work...
     
  18. Apr 17, 2005 #17

    aek

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    Can the attitude?
    Let me first explain this to you, i've posted this 3 days ago, and yet i haven't go a decent answer. All u've guys have done is go around circles without touching the question. I used the product rule to its perfection. THATS MY WORKING BUDDY. If you think your some FBI trying to find out if i want you to the my homework then YOUR wrong. If you don't want to help with my question THEN don't post and waste my time. WHEN i help someone on this forum, i give 100%. Now how about you CAN your apathy and actually try to be a mentor like your supposed to be. BYE
     
  19. Apr 17, 2005 #18

    Hurkyl

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    Yes, can the attitude.

    Let me explain this to you: first, I'm sure you've noticed we have guidelines for homework help. To quote a specific line from them:

    "So, post away--and show your work!"

    Saying "I used the product rule to its perfection" is not showing your work.


    Furthermore, I don't understand how you can be so utterly confident you did the problem perfectly that you absolutely refuse to show what you've done (so that, you know, we could find your mistake), and at the same time be so insolent when someone suggests that the answer to which you are trying to compare is wrong.

    Either you're wrong (in which case you need to show your work), or the book's wrong. Actually, both could be wrong... but you seem to accept neither possibility.
     
  20. Apr 17, 2005 #19

    aek

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    As far as i'm concerned i'm on the lines with the guidelines. Must i remind you Hurkyl, this question isn't based upon the power rule alone. I know for a fact that i did the power rule correct but simplifying and getting neat is where i find trouble. And i asked did anyone else get an answer similar to the book. Honestly, Hurkyl, did you even look at the question? Once again Hurkyl, don't waste my time because i know for a fact you have no intention of helping people at all and im pretty sure your position would of been put under review if someone else can spare soo much time on this forum (that time is indeed a waste).
     
  21. Apr 17, 2005 #20

    dextercioby

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    What does the problem ask?Does it ask to simply differentiate the function [itex] x\sqrt{x-1} [/itex],or it asks for something totally different.

    Daniel.
     
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