Results 1 to 8 of 8

Math Help - finding derivative

  1. #1
    rmn
    rmn is offline
    Junior Member
    Joined
    Sep 2007
    Posts
    26

    finding derivative

    y= ((2x-5)^-1 ) * (x^(2)-5x)^6

    this is what i have so far

    dy/dx=

    (2x-5)^-1 * 6(x^2-5x)^5 * (2x-5) * 2 +

    (x^2-5x)^6 * -1(2x-5)^-2 * 2

    cancel the (2x-5) in the first term right

    so then i have

    12(x^2-5x)^5- 2(x^2-5x)^6 * (2x-5)^-2

    factor out 2(x^2-5x)^5

    so i have

    2(x^2-5x)^5 [ 6-(x^2-5x)*(2-5)^-2]

    now i am confused how to simplify this
    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 rmn View Post
    y= ((2x-5)^-1 ) * (x^(2)-5x)^6

    this is what i have so far

    dy/dx=

    (2x-5)^-1 * 6(x^2-5x)^5 * (2x-5) * 2 +

    (x^2-5x)^6 * -1(2x-5)^-2 * 2
    what is that 2 for?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    rmn
    rmn is offline
    Junior Member
    Joined
    Sep 2007
    Posts
    26
    Quote Originally Posted by Jhevon View Post
    what is that 2 for?
    derivative of the 2x-5
    yeah i wasn't sure if i was supposed to do that?
    is it part of the chain rule or something
    Follow Math Help Forum on Facebook and Google+

  4. #4
    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 rmn View Post
    derivative of the 2x-5
    yeah i wasn't sure if i was supposed to do that?
    is it part of the chain rule or something
    no. the chain rule was done when you wrote (2x - 5). you don't need to take the derivative of that.

    so you should have: (2x-5)^{-1}(6) \left( x^2-5x \right)^5(2x-5) + (x^2-5x)^6 (-1)(2x-5)^{-2}(2)

    now continue. what do we have common here? what can we factor out? and what powers should we factor out?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    rmn
    rmn is offline
    Junior Member
    Joined
    Sep 2007
    Posts
    26
    Quote Originally Posted by Jhevon View Post
    no. the chain rule was done when you wrote (2x - 5). you don't need to take the derivative of that.

    so you should have: (2x-5)^{-1}(6) \left( x^2-5x \right)^5(2x-5) + (x^2-5x)^6 (-1)(2x-5)^{-2}(2)

    now continue. what do we have common here? what can we factor out? and what powers should we factor out?
    oh, just a general question how do we know when to stop using the chain rule like in that question
    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 rmn View Post
    oh, just a general question how do we know when to stop using the chain rule like in that question
    the chain rule says: \frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x) ........we take the derivative of the inside function once. we don't take the derivative of the derivative of g(x) (that would be g''(x), and that's not in the formula). (the only time we may have to do that is if g(x) itself is a composite function, which was not the case here)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    rmn
    rmn is offline
    Junior Member
    Joined
    Sep 2007
    Posts
    26
    Quote Originally Posted by Jhevon View Post
    the chain rule says: \frac d{dx}f(g(x)) = f'(g(x)) \cdot g'(x) ........we take the derivative of the inside function once. we don't take the derivative of the derivative of g(x) (that would be g''(x), and that's not in the formula). (the only time we may have to do that is if g(x) itself is a composite function, which was not the case here)

    oh ok thanks
    ok so now i have

    2(x^2-5x)^5 [ 3- (x^2-5x)(2x-5)^-2]

    is that right
    Follow Math Help Forum on Facebook and Google+

  8. #8
    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 rmn View Post
    oh ok thanks
    ok so now i have

    2(x^2-5x)^5 [ 3- (x^2-5x)(2x-5)^-2]

    is that right
    yes, that's fine
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding a derivative.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 21st 2011, 02:37 PM
  2. Help finding derivative
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 29th 2010, 09:00 PM
  3. Replies: 1
    Last Post: January 21st 2010, 12:08 PM
  4. finding the derivative
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 24th 2009, 09:12 AM
  5. Finding the third derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 18th 2009, 08:43 PM

Search Tags


/mathhelpforum @mathhelpforum