Results 1 to 8 of 8

Math Help - Determing the derivative of a function using the chain rule

  1. #1
    !!!
    !!! is offline
    Junior Member
    Joined
    Feb 2008
    Posts
    28

    Determing the derivative of a function using the chain rule

    How do I get from the problem to the answer using the chain rule?
    It would be much appreciated if someone indicated the steps for me, thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by !!! View Post
    How do I get from the problem to the answer using the chain rule?
    It would be much appreciated if someone indicated the steps for me, thank you!
    To differentiate the product you use the product rule. As part of that process you need to differentiate (x^2 + 2)^{1/3}. Get that derivative using the chain rule.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    Only one comment to add: you recently posted the same problem on MathLinks. They don't actually reply as faster as we can, so I suggest you to stay here 'cause you'll get a quickly answer and full solutions. (This last thing depends of your problem.)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    !!!
    !!! is offline
    Junior Member
    Joined
    Feb 2008
    Posts
    28
    Alright, thanks for the help and heads-up, guys.
    I got through the part where I already performed the product rule and chain rule, but so far, my answer looks like this:


    How do I get from that to the correct answer of
    ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by !!! View Post
    Alright, thanks for the help and heads-up, guys.
    I got through the part where I already performed the product rule and chain rule, but so far, my answer looks like this:


    How do I get from that to the correct answer of
    ?
    Note that (x^2 + 2)^{-2/3} = (x^2 + 2)^{1/3} (x^2 + 2)^{-1} = \frac{(x^2 + 2)^{1/3}}{(x^2 + 2)}. Make that substitution and then factorise by taking out the common factor of 2x (x^2 + 2)^{1/3} in each term.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by !!! View Post
    Alright, thanks for the help and heads-up, guys.
    I got through the part where I already performed the product rule and chain rule, but so far, my answer looks like this:


    How do I get from that to the correct answer of
    ?
    y'=2x(x^2+2)^\frac{1}{3}+(x^2+1)\cdot\frac{1}{3}(x  ^2+2)^\frac{-2}{3}(2x)

    rewriting the second term with a positive exponent gives

    y'=2x(x^2+2)^\frac{1}{3}+(2x)(x^2+1)\cdot\frac{1}{  3(x^2+2)^\frac{2}{3}}

    multiplying the numerator and denominator of the second term

    [tex]MATH]

    simplifying

    y'=2x(x^2+2)^\frac{1}{3}+(2x)(x^2+1)\frac{(x^2+2)^  \frac{1}{3}}{3(x^2+2)}

    factoring out the GCF gives

    y'=2x(x^2+2)^\frac{1}{3}\left[1+\frac{x^2+1}{3(x^2+2)}\right]
    Follow Math Help Forum on Facebook and Google+

  7. #7
    !!!
    !!! is offline
    Junior Member
    Joined
    Feb 2008
    Posts
    28
    Thank you very much, everyone. Now I understand how to do that problem.
    I also have trouble determining the derivatives of these and I'm not sure if I'm doing them correctly because the book don't have their answers.

    1. y = x sin x^{1/2}
    y' = x cos (x^{1/2}) d/dx (x^{1/2}) + (sin x^{1/2}) d/dx (x^{1/2})
    y' = x cos \sqrt{x} + sin  \sqrt{x} / {2x \sqrt{x}}

    2. y = x/(7-3x^{1/2})
    I only got up to (7-3x)^{1/2} (-3) - x[(1/2)(7-3x^{1/2})] (-3)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by !!! View Post
    Thank you very much, everyone. Now I understand how to do that problem.
    I also have trouble determining the derivatives of these and I'm not sure if I'm doing them correctly because the book don't have their answers.

    1. y = x sin x^{1/2}
    y' = x cos (x^{1/2}) d/dx (x^{1/2}) + (sin x^{1/2}) d/dx (x^{1/2})
    y' = x cos \sqrt{x} + sin  \sqrt{x} / {2x \sqrt{x}}

    2. y = x/(7-3x^{1/2})
    I only got up to (7-3x)^{1/2} (-3) - x[(1/2)(7-3x^{1/2})] (-3)
    1. You need the product rule:

    u = x \Rightarrow \frac{du}{dx} = 1

    v = \sin x^{1/2} \Rightarrow \frac{dv}{dx} = \cos x^{1/2} \times \frac{1}{2} x^{-1/2} = \cos \sqrt{x} \times \frac{1}{2\sqrt{x}}, where the chain rule has also been used.

    Then you substitute the above results into \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}.

    Some simplification of the resulting answer will be possible .....

    ----------------------------------------------------------------------------------------------

    2. You need the quotient rule:

    u = x \Rightarrow \frac{du}{dx} = 1

    v = 7 - 3 x^{1/2} \Rightarrow \frac{dv}{dx} = -\frac{3}{2} x^{-1/2} = -\frac{3}{2\sqrt{x}}.

    Then you substitute the above results into \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}.

    Some simplification (requiring a little bit of algebra) of the resulting answer will be desirable .....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] find derivative of a trig function (chain rule included)
    Posted in the Calculus Forum
    Replies: 10
    Last Post: May 8th 2011, 05:20 PM
  2. Replies: 2
    Last Post: August 14th 2010, 05:45 AM
  3. Replies: 5
    Last Post: October 19th 2009, 01:04 PM
  4. Derivative and Chain Rule HELP!!!!
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 31st 2008, 06:06 AM
  5. Replies: 2
    Last Post: December 13th 2007, 05:14 AM

Search Tags


/mathhelpforum @mathhelpforum