Results 1 to 3 of 3

Math Help - chain rule help

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    19

    chain rule help

    can please someone help me how to solve for the derivative of these

    y = [t^2 + (1/t)]^2

    G(x) = {1+[x+(x^2+x^3)^4]^5}^6

    h(x) = {x+[x+(x)^1/2]^1/2}^1/2

    y = x^1/2 sin [x+(x)^1/2]^1/2 cot x^1/2


    if possible, not simplified. thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    y = (t^2 + (1/t))^2, firstly rewrite this in the form y = (t^2 + t^{-1})^2).

    To apply the chain rule you take the function inside the brackets to the front, like so:

    = 2t - t^{-2} . (t^2 + t^{-1})^2

    You then multiply the whole thing by the power of the bracket:

    = 2t - t^{-2} . (t^2 + t^{-1})^{2} . 2

    Then finally you reduce the power of the bracket by one, ending up with:

    \frac{dy}{dx}= 2t - t^{-2} . (t^2 + t^{-1})^{1} . 2

    \frac{dy}{dx} = 2(2t - t^{-2})(t^2 + t^{-1})

    Im sure you can simply the remaining equation and apply the same rules to your other questions
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,391
    Thanks
    55
    Quote Originally Posted by craig View Post
    y = (t^2 + (1/t))^2, firstly rewrite this in the form y = (t^2 + t^{-1})^2).

    To apply the chain rule you take the function inside the brackets to the front, like so:

    = 2t - t^{-2} . (t^2 + t^{-1})^2

    You then multiply the whole thing by the power of the bracket:

    = 2t - t^{-2} . (t^2 + t^{-1})^{2} . 2

    Then finally you reduce the power of the bracket by one, ending up with:

    \frac{dy}{dx}= 2t - t^{-2} . (t^2 + t^{-1})^{1} . 2

    \frac{dy}{dx} = 2(2t - t^{-2})(t^2 + t^{-1})

    Im sure you can simply the remaining equation and apply the same rules to your other questions
    The second and third steps are incorrect. Here's a way that always work.

    First let me say that the chain rule is used for a composition of functions. In this case, the order of operation is

    (i) t^2 + \frac{1}{t}

    (ii) then square the result from (i) i.e. \left( t^2 + \frac{1}{t} \right)^2

    For the chain rule, let

    u = t^2 + \frac{1}{t}

    so

    y = u^2

    Then the derivative \frac{dy}{dt} is

    \frac{dy}{dt} = \frac{dy}{du} . \frac{du}{dt}

    so \frac{dy}{du} = 2u and \frac{du}{dt} = 2t - \frac{1}{t^2}
    so
    \frac{dy}{dt} = \frac{dy}{du} . \frac{du}{dt} = 2u \left( 2t - \frac{1}{t^2} \right) = 2 \left( t^2 + \frac{1}{t} \right) \left( 2t - \frac{1}{t^2} \right)

    It is the answer given above but I think it's necessary to put in these steps until you get the hang of it. Once you do, you can write the answer in one step (working from out to in on your function).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: November 9th 2010, 02:40 AM
  2. Chain Rule Inside of Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 22nd 2009, 09:50 PM
  3. Replies: 5
    Last Post: October 19th 2009, 02:04 PM
  4. Replies: 3
    Last Post: May 25th 2009, 07:15 AM
  5. Replies: 2
    Last Post: December 13th 2007, 06:14 AM

Search Tags


/mathhelpforum @mathhelpforum