Results 1 to 5 of 5

Math Help - Calculus : Chain Rule help

  1. #1
    Senior Member
    Joined
    Jul 2007
    Posts
    290

    Calculus : Chain Rule help

    I'm asked to find the derivative of the function.

    y = R / ( R^2 + 1)^(1/2)


    Now I know i'm gonna have to use the quotient rule and incorporate the chain rule in the midst of it. My whole problem is simplifying afterwards. Will someone help me? Thanks

    This is what I've got :

    ((r^2 + 1)^1/2 * 1 - ( r * (1/2)(R^2 + 1)^(-1/2) * 2r)) / r^2 + 1


    How do I simplify this so it'll end up looking like

    y' = (r^2 + 1)^(-3/2)


    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Edit: I didnt figure it out.... any help is appreciated
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    768
    Hello, Jonathan!

    Differentiate: . y \:= \:\frac{r}{(r^2 + 1)^{\frac{1}{2}}}

    My problem is simplifying.

    This is what I've got: . y' \;=\;\frac{(r^2 + 1)^{\frac{1}{2}}\cdot1 -  r\cdot\frac{1}{2}(r^2 + 1)^{-\frac{1}{2}}\cdot2r}{r^2 + 1} . . . . Correct!

    How do I simplify this so it'll end up looking like: . y' \;= \;(r^2 + 1)^{-\frac{3}{2}}
    \text{You have: }\;y' \;=\;\frac{(r^2+1)^{\frac{1}{2}} - r^2(r^2+1)^{-\frac{1}{2}}}{r^2+1}


    Mutiply top and bottom by (r^2+1)^{\frac{1}{2}}

    . . y' \;=\;\frac{(r^2+1)^{\frac{1}{2}}}{(r^2+1)^{\frac{1  }{2}}} \cdot \frac{(r^2+1)^{\frac{1}{2}} - r^2(r^2+1)^{-\frac{1}{2}}} {r^2+1} \;=\;\frac{(r^2+1) - r^2}{(r^2+1)^{\frac{3}{2}}} \;=\;\frac{1}{(r^2+1)^{\frac{3}{2}}}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Quote Originally Posted by Soroban View Post
    Hello, Jonathan!

    \text{You have: }\;y' \;=\;\frac{(r^2+1)^{\frac{1}{2}} - r^2(r^2+1)^{-\frac{1}{2}}}{r^2+1}


    Mutiply top and bottom by (r^2+1)^{\frac{1}{2}}

    . . y' \;=\;\frac{(r^2+1)^{\frac{1}{2}}}{(r^2+1)^{\frac{1  }{2}}} \cdot \frac{(r^2+1)^{\frac{1}{2}} - r^2(r^2+1)^{-\frac{1}{2}}} {r^2+1} \;=\;\frac{(r^2+1) - r^2}{(r^2+1)^{\frac{3}{2}}} \;=\;\frac{1}{(r^2+1)^{\frac{3}{2}}}


    Doh! Question, why multiply by (r^2 +1 ) ^1/2 ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    Quote Originally Posted by JonathanEyoon View Post
    Doh! Question, why multiply by (r^2 +1 ) ^1/2 ?
    Because it will rationalize the numerator:
    = (r^2+1)^{\frac 12+\frac 12}- r^2(r^2+1)^{\frac 12-\frac 12}

    = (r^2+1)^1- r^2(r^2+1)^0

    = r^2+1- r^2

    = +1

    The thing you should notice is that both terms in the numerator have powers of (r^2+1)^{\frac 12} in them. Which upon closer analysis, reveals that you can make the numerator rational by multiplying by (r^2+1)^{\frac 12}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Calculus chain rule questions
    Posted in the Calculus Forum
    Replies: 8
    Last Post: May 10th 2010, 09:51 PM
  2. Multivariable Calculus - Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 16th 2010, 06:11 PM
  3. Calculus chain rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 27th 2009, 05:20 PM
  4. calculus: the chain rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 23rd 2009, 01:01 PM
  5. Calculus - Chain Rule Derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 10th 2007, 06:05 PM

Search Tags


/mathhelpforum @mathhelpforum