Results 1 to 9 of 9

Math Help - Why are they using the chain rule on this question?

  1. #1
    Junior Member NAPA55's Avatar
    Joined
    Mar 2008
    Posts
    58

    Why are they using the chain rule on this question?

    I'm stuck on a question from my chapter review homework. It appears the textbook has used the Chain Rule, but we were never taught how to use the Chain Rule in this context, so I don't know how to apply it.

    How would I work this question out?

    Find the derivative.
     y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2}

    I tried to chain the top and then the bottom separately- obviously that can't be right.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by NAPA55 View Post
    I'm stuck on a question from my chapter review homework. It appears the textbook has used the Chain Rule, but we were never taught how to use the Chain Rule in this context, so I don't know how to apply it.

    How would I work this question out?

    Find the derivative.
     y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2}

    I tried to chain the top and then the bottom separately- obviously that can't be right.
    we use the chain rule to find the derivative of composite functions. clearly, (x - 4)^5 and (2x^3 - 1)^2 are composite functions. you use the chain rule when differentiating them. otherwise, you would use the quotient and product rules to differentiated the whole function
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member NAPA55's Avatar
    Joined
    Mar 2008
    Posts
    58
    This is what I tried to do:

    I tried to chain the top to find the derivative of f(x), which is the top function.

    I tried to chain the bottom to find the derivative of g(x), which is the bottom function.

    I got this, and I tried to use the quotient rule after I used the chain rule.
    f(x) = (x+2)(x-4)^5 f '(x) = (x-4)^5 + 5(x+2)(x-4)^4

    g(x) = (2x^3-1)^2 g '(x) = 2(2x^3-1)(6x) = 12x(2x^3-1)

    But the book chained the whole function. They threw the whole original function out front, as you would do with the chain rule. This is their response:
    y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2} [\frac{1}{x+2} + \frac{5}{x-4} - \frac{12x^2}{2x^3-1}]
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    logarithmic differentiation

    It would be a combination of rules chain quotent ect.

    Why not try logarithmic differentiation
    <br />
y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2}<br />


    <br />
ln(y) = ln \left(\frac{(x+2)(x-4)^5}{(2x^3-1)^2} \right)=ln(x+2)+5ln(x-4)-2ln(2x^3-1)<br />

    taking the derivative

    \frac{1}{y}\frac{dy}{dx}=\frac{1}{(x+2)}+\frac{5}{  (x-4)}-\frac{12x^2}{(2x^3-1)}

    Solving for \frac{dy}{dx} gives

    \frac{dy}{dx}=y \left( \frac{1}{(x+2)}+\frac{5}{(x-4)}-\frac{12x^2}{(2x^3-1)}\right)=

    \frac{dy}{dx}= \frac{(x+2)(x-4)^5}{(2x^3-1)^2} \cdot \left( \frac{1}{(x+2)}+\frac{5}{(x-4)}-\frac{12x^2}{(2x^3-1)}\right)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member NAPA55's Avatar
    Joined
    Mar 2008
    Posts
    58
    If logarithmic differentiation is needed to do that question, we skipped that topic in this unit, and as far as I know, we're not responsible for it.

    If this is a logarithmic differentiation type question that was on the review, I guess I shouldn't have attempted it yet. This always happens to me! I try a question, get frustrated because I can't do it, and then later learn it wasn't something we were taught yet! LOL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by NAPA55 View Post
    This is what I tried to do:

    I tried to chain the top to find the derivative of f(x), which is the top function.

    I tried to chain the bottom to find the derivative of g(x), which is the bottom function.

    I got this, and I tried to use the quotient rule after I used the chain rule.
    f(x) = (x+2)(x-4)^5 f '(x) = (x-4)^5 + 5(x+2)(x-4)^4

    g(x) = (2x^3-1)^2 g '(x) = 2(2x^3-1)(6x) = 12x(2x^3-1)

    But the book chained the whole function. They threw the whole original function out front, as you would do with the chain rule. This is their response:
    y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2} [\frac{1}{x+2} + \frac{5}{x-4} - \frac{12x^2}{2x^3-1}]
    your derivative for the second is not correct, you should have 6x^2 not 6x. maybe that's why you have been having problems.


    Quote Originally Posted by NAPA55 View Post
    If logarithmic differentiation is needed to do that question, we skipped that topic in this unit, and as far as I know, we're not responsible for it.

    If this is a logarithmic differentiation type question that was on the review, I guess I shouldn't have attempted it yet. This always happens to me! I try a question, get frustrated because I can't do it, and then later learn it wasn't something we were taught yet! LOL
    logarithmic differentiation is not REQUIRED, it is just another way to do the problem. this problem can be done with only the chain, quotient and product rules.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member NAPA55's Avatar
    Joined
    Mar 2008
    Posts
    58
    That was a typo on my part. It should be 12x^2(2x^3-1). Should it work if I keep going from what I had?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by NAPA55 View Post

    But the book chained the whole function. They threw the whole original function out front, as you would do with the chain rule. This is their response:
    y = \frac{(x+2)(x-4)^5}{(2x^3-1)^2} [\frac{1}{x+2} + \frac{5}{x-4} - \frac{12x^2}{2x^3-1}]
    Because of the form of the book answer I would guess "they" (The book )used logarithmic differentiation

    If you have time to learn it, it is a great tool for taking derivatives.

    Good luck.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by NAPA55 View Post
    That was a typo on my part. It should be 12x^2(2x^3-1). Should it work if I keep going from what I had?
    yes, keep going.

    TheEmptySet has a good point. it would seem that the book did use logarithmic differentiation. it can be done without it though, so continue. i also agree with TheEmptySet in that, if you can learn logarithmic differentiation by yourself, you should. you are familiar with the laws of logarithms right?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Chain rule question #3
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 7th 2010, 08:07 AM
  2. Another chain rule question.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 7th 2010, 05:53 AM
  3. [SOLVED] Another Chain Rule Question.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 13th 2010, 04:31 PM
  4. Replies: 5
    Last Post: October 19th 2009, 02:04 PM
  5. chain rule question
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 6th 2008, 10:48 AM

Search Tags


/mathhelpforum @mathhelpforum