Results 1 to 9 of 9

Math Help - Stuck on derivative.

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    149

    Stuck on derivative.

    I have equation


    y=\frac{2x-5}{3x+2}

    Please check my working.(I started expanding the right hand side of the numerator first , i hope you can follow.

    =\frac{[(3x+2)^2(2)]-[(2x-5)[(3x+2)(3x+2)]}{[(3x+2)^2]^2}


    =\frac{[(3x+2)^2(2)]-[(2x-5)(9x^2+6x+6x+4)]}{[3x+2)^2]^2}

    =\frac{[(3x+2)^2(2)]-[(2x-5)(9x^2+12x)]}{[(3x+2)^2]^2}

    =\frac{[(3x+2)^2(2)]-[(2x-5)(18x+12)]}{[(3x+2)^2]^2}


    =\frac{[((3x+2)(3x+2)(2)]-(36x+24x-90x-60)}{[(3x+2)(3x+2)(3x+2)(3x+2)]}

    =\frac{((3x+2)(3x+2)(2)-36x-24x+90x+60} {(3x+2)(3x+2)(3x+2)(3x+2)}

    =\frac{(9x^2+12x+4)(2)+30x+60}{54x+78x+24}
    =\frac{(18x^2+24x+8)+30x+60}{132x+24}
    =\frac{18x^2+54x+68}{132x+24}

    ok i know thats alot to read but if anyone is able to let me know if ive gone wrong anywhere it would be much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by el123 View Post

    [snip]

    I have equation


    y=\frac{2x-5}{3x+2}

    Please check my working.(I started expanding the right hand side of the numerator first , i hope you can follow.

    =\frac{[(3x+2)^{{\color{red}2}}(2)]-[(2x-5)[{\color{blue}(3x+2)}{\color{red}(3x+2)}]}{[(3x+2)^2]^{{\color{red}2}}}


    [/snip]
    The terms in red aren't needed, and the term in blue needs to be differentiated. That will give you the answer you're looking for.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    I would divide first to make the calculus a lot easier.
    That way you can avoid the quotient rule.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2009
    Posts
    149
    Darn i just realised the very first equation was

    \frac{2x-5}{(3x+2)^2}

    does it make sense now or still not right?

    Woulld you be able to explain the steps of how i should begin to diffrentiate it?
    Last edited by mr fantastic; August 12th 2009 at 04:47 AM. Reason: m --> r
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,628
    Thanks
    430
    Quote Originally Posted by el123 View Post
    Damn i just realised the very first equation was

    \frac{2x-5}{(3x+2)^2}

    does it make sense now or still not right?

    Woulld you be able to explain the steps of how i should begin to diffrentiate it?
    seems like I've seen this question before ...

    \frac{d}{dx}\left[\frac{2x-5}{(3x+2)^2}\right]

    \frac{(3x+2)^2(2) - (2x-5) \cdot 6(3x+2)}{(3x+2)^4}

    \frac{34-6x}{(3x+2)^3}

    you need to learn about the chain rule for derivatives so that you may easily find derivatives of functions like (3x+2)^2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by el123 View Post
    Darn i just realised the very first equation was

    \frac{2x-5}{(3x+2)^2}

    does it make sense now or still not right?

    Woulld you be able to explain the steps of how i should begin to diffrentiate it?
    Logarithmic differentiation can also be used: Logarithmic Differentiation
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jul 2009
    Posts
    149
    Quote Originally Posted by skeeter View Post
    seems like I've seen this question before ...

    \frac{d}{dx}\left[\frac{2x-5}{(3x+2)^2}\right]

    \frac{(3x+2)^2(2) - (2x-5) \cdot 6(3x+2)}{(3x+2)^4}

    \frac{34-6x}{(3x+2)^3}

    you need to learn about the chain rule for derivatives so that you may easily find derivatives of functions like (3x+2)^2
    ok i can see how you get the last part of the numerator and ive been learning the chain rul ebut i cant seem to see how you expanded that to get 34-6x/3x+2^3

    Could you explain please. I am so confused. I keep getting answers that are way off the mark
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,628
    Thanks
    430
    note the common factor (3x+2) in both terms of the numerator ...

    \frac{\textcolor{red}{(3x+2)^2}(2) - (2x-5) \cdot 6\textcolor{red}{(3x+2)}}{(3x+2)^4}

    factor out one (3x+2) from both terms ...

    \frac{\textcolor{red}{(3x+2)}[(3x+2)(2) - (2x-5) \cdot 6]}{(3x+2)^4}

    cancel the common factor with one in the denominator ...

    \frac{(3x+2)(2) - (2x-5) \cdot 6}{(3x+2)^3}

    clean up the like terms ...

    \frac{6x+4 - 12x+30}{(3x+2)^3}

    \frac{34-6x}{(3x+2)^3}

    the hardest thing about calculus is the algebra.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    ... perhaps matheagle is right to suggest avoiding the quotient rule, which often raises the power in the denominator unnecessarily. Another way of avoiding it is to use a chain rule...



    inside a product rule...



    giving...



    or




    ...where the simplest common denominator is formed directly.


    _______________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus Forum
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stuck on part 2 on this derivative question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 21st 2011, 05:58 PM
  2. [SOLVED] Partial derivative. Completely stuck
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 18th 2010, 02:40 AM
  3. I'm stuck on this derivative...
    Posted in the Calculus Forum
    Replies: 10
    Last Post: March 25th 2010, 03:16 PM
  4. A little stuck on a derivative
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 8th 2007, 09:08 AM
  5. Stuck, Stuck, Stuck - Need Help Urgently
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 17th 2005, 05:46 AM

Search Tags


/mathhelpforum @mathhelpforum