1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Chain rule in division rule

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data

    y = (2x-3)/(x^2+4)^2

    2. Relevant equations



    3. The attempt at a solution

    I am trying to relearn the calculus that I forgot from many moons ago. I am struggling with the chain rule in the above example. I tried to set it up as follows:

    This is what I know u=x^2+4 u'=2x

    [(x^2+4)^2*Dx (2x-3)-(2x-3) ?? ]/(x^2+4)^4

    I am confused when it comes to setting up the second half. Any help would be much obliged. I just can't seem to understand how to set up the problem


    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 6, 2007 #2
    [tex]y=\frac{2x-3}{(x^2+4)^2}[/tex]
    [tex]\frac{dy}{dx}=\frac{2(x^2+4)^2-(2x-3)(2(x^2+4)(2x))}{(x^2+4)^4}[/tex]
    [tex]=\frac{2(x^2+4)^2-(4x^2-12x)(x^2+4)}{(x^2+4)^4}[/tex]
    [tex]=\frac{2(x^2+4)-(4x^2-12x)}{(x^2+4)^3}[/tex]

    How do I make a new line? \\ doesn't seem to work.
     
    Last edited: May 6, 2007
  4. May 6, 2007 #3

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I would imagine just breaking up the tex tags works

    [tex]y=\frac{2x-3}{(x^2+4)^2}[/tex]

    [tex]\frac{dy}{dx}=\frac{2(x^2+4)^2-(2x-3)(2(x^2+4)(2x))}{(x^2+4)^4}[/tex]

    [tex]=\frac{2(x^2+4)^2-(4x^2-12x)(x^2+4)}{(x^2+4)^4}[/tex]

    [tex]=\frac{2(x^2+4)-(4x^2-12x)}{(x^2+4)^3}[/tex]
     
  5. May 7, 2007 #4
    Thanks for the reply with setting up the Calculus part. Now I know that my algebra is a little rusty, but is there a mistake in the change from the following two lines? Shouldn't it be
    (8x^2-12x)(x^2+4) not (4x^2-12x)(x^2_4)


    [tex]=\frac{2(x^2+4)^2-(4x^2-12x)(x^2+4)}{(x^2+4)^4}[/tex]

    [tex]=\frac{2(x^2+4)-(4x^2-12x)}{(x^2+4)^3}[/tex]
     
  6. May 7, 2007 #5
    Yes it should be 8x^2-12x.
     
  7. May 7, 2007 #6
    Bad algebra, sorry.
     
  8. May 7, 2007 #7
    well my algebra is very rusty to say the least... I have been out of math class a few years, and am working towards going back. If I was sure I was right I wouldn't have asked if it was wrong. Thanks again for all the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Chain rule in division rule
  1. Chain rule (Replies: 1)

  2. Chain rule (Replies: 2)

  3. Chain rule (Replies: 22)

Loading...