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Quotient Rule

  1. May 10, 2013 #1
    1. The problem statement, all variables and given/known data
    f(x) is given by the forumula [itex]y=\sqrt{3x^2 + 2x + 1}[/itex]

    Find A: The first derivative
    B: The second derivative


    2. Relevant equations
    chain rule
    quotient and product rule?


    3. The attempt at a solution
    I think I have made a good logical attempt at part A but only have an inclin when it comes to part B so anyway here is my part A:

    [tex]
    y=(3x^2 + 2x + 1)^{\frac{1}{2}} \\
    u = 3x^2 + 2x + 1 \\
    u\prime = 6x + 2 \\
    y\prime (u) = \frac{1}{2} u^{-\frac{1}{2}} \\
    y\prime= \frac{1}{2} u^{-\frac{1}{2}} \times 6x + 2 \\
    y\prime = \frac{1}{2}(3x^2 + 2x + 1)^{-\frac{1}{2}} \times 6x + 2 \\
    y\prime = (3x + 1)(3x^2 + 2x + 1)^{-\frac{1}{2}} \\
    y\prime = \frac{3x + 1}{\sqrt{3x^2 + 2x + 1}} \\
    [/tex]

    For part B I am unsure as to where to start, if I convert the denominator of the fraction (to my final answer) to a power of a half then I know already the derivative of it as found above or if I leave it as a product of minus a half then could I simply apply the product rule to it?

    EDIT: I am unsure as to why the latex code is not rendering properly, If anyone can spot a mistake in the code I would be very grateful :)
     
    Last edited: May 10, 2013
  2. jcsd
  3. May 10, 2013 #2
    The problem with your code is probably with either mismatched or broken tex tags. I get that all the time when I mismatch a tex with an itex.
     
  4. May 10, 2013 #3
    You didn't close the last fraction after closing the sqrt.
     
  5. May 10, 2013 #4
    Thank you! I have fixed it now :)
     
  6. May 10, 2013 #5
    I always make that mistake :) now I'm a fraction closure checking machine!
     
  7. May 10, 2013 #6
    Oh for crying out loud. I didn't understand how you got that derivative and I'm sitting here typing out hints for you to integrate the derivative before I realized how dyslexic I was being.. >_< personally I vote for the product rule. I always vote for the product rule. On the other hand, when you square the denominator you get rid of that square root. Either way it will be a little bit messy.
     
  8. May 10, 2013 #7
    :), Thanks. I just realised that the title to this thread should be the chain rule! I guess I had the quotient rule in my head as I was creating the thread as I was thinking about finding the derivative of my final answer to Part A. I have had a go using the quotient rule and yes it is really really messy! I will try the product rule now and see what happens.

    This might be a stupid question but when using the quotient rule, for example on my final answer to Part A, would v' be my final answer to Part A? or will it be to the power minus half as it is on the denominator of the fraction?
     
  9. May 11, 2013 #8
    Anyway here is my attempy at Part B, using the quotient rule on my final answer to part B.

    [tex]
    y\prime\prime = \frac{((3x^2 + 2x + 1)^{\frac{1}{2}})3)-((3x+1)(3x+1)(3x^2 + 2x + 1)^{-\frac{1}{2}}))}{3x^2 + 2x + 1} \\
    y\prime\prime = \frac{(3\sqrt{3x^2 + 2x + 1} - \frac{(3x+1)^2}{\sqrt{3x^2 + 2x + 1}}}{3x^2 + 2x + 1}
    [/tex]
    If anyone could check it please :)
     
  10. May 11, 2013 #9

    Dick

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    It's correct, but you could simplify it a lot.
     
  11. May 11, 2013 #10
    Thanks, any tips on simplifying this monster would be appreciated :)
     
  12. May 11, 2013 #11

    Dick

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    Put the terms in the numerator over a common denominator. The sqrt part.
     
  13. May 13, 2013 #12
    my tutors has marked it as is and then told me the simplest way of writing it is as below,

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

    Any help showing how to get from mine to this would be appreciated, I am a bit lost now!

    Thanks,
     
  14. May 13, 2013 #13

    micromass

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    Did you follow Dick his suggestion? What do you get?
     
  15. May 13, 2013 #14
    Didn't really no where to start to be honest. I cant seem to manipulate it any more. Ive been out of education since I was 16, (am now 24) and to start full time uni in September embarked upon a fast track course this academic year ending now, its all well and good for teaching things we need to know like calculus et al, but falls short on the lower level stuff that would have been taught when I were 16. I feel kind of embarrassed to be honest!
     
  16. May 13, 2013 #15

    micromass

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    If you recall operations on fractions, then you'll see that

    [tex]a + \frac{b}{c} = \frac{ac}{c} + \frac{b}{c} = \frac{ac + b}{c}[/tex]

    The numerator of your result is in the above form, so apply the simplification.
     
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