Results 1 to 7 of 7

Math Help - Simple Derivative

  1. #1
    Newbie
    Joined
    Jul 2007
    Posts
    4

    Simple Derivative

    Hello,

    I'm a bit rusty on my calculus and I have this function and when I use the chain rule I get the f'(x) function



    I must be doing something wrong though as when x = 9.17, the dy/dx value should be about 16 but it ends up being about .33 something. Can anyone point out where I went wrong? Thanks in advance.
    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 Jimpson View Post
    Hello,

    I'm a bit rusty on my calculus and I have this function and when I use the chain rule I get the f'(x) function



    I must be doing something wrong though as when x = 9.17, the dy/dx value should be about 16 but it ends up being about .33 something. Can anyone point out where I went wrong? Thanks in advance.
    how did you get 75 in the top?

    obviously you're using the quotient rule incorrectly.

    recall:

    Quotient Rule:

    \left( \frac {u}{v} \right)' = \frac {vu' - uv'}{v^2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2007
    Posts
    4
    Quote Originally Posted by Jhevon View Post
    how did you get 75 in the top?
    150 * (1+49.1e^-.5x)^-1

    -150 * (1+49.1e^.5x)^-2 * -.5e^.5x

    75e-.5x
    (1+49.1e^.5x)^2

    If that makes sense.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2007
    Posts
    4
    Quote Originally Posted by Jhevon View Post
    how did you get 75 in the top?

    obviously you're using the quotient rule incorrectly.

    recall:

    Quotient Rule:

    \left( \frac {u}{v} \right)' = \frac {vu' - uv'}{v^2}
    I thought quotient rules were used for a function on both the top and bottom of the fraction?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    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 Jimpson View Post
    150 * (1+49.1e^-.5x)^-1

    -150 * (1+49.1e^.5x)^-2 * -.5e^.5x

    75e-.5x
    (1+49.1e^.5x)^2

    If that makes sense.
    ok, so you used the product rule. but note that you have the wrong derivative for the inside.

    the derivative of 1 + 49.1e^{-0.5x} is -24.55e^{-0.5x}, there's your problem.

    remember, you have a constant 49.1 multiplying the e
    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 Jimpson View Post
    I thought quotient rules were used for a function on both the top and bottom of the fraction?
    well, technically a constant is a function. it will work anyway. you can use the product rule if you wish, i gave you the corrections already
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jul 2007
    Posts
    4
    Quote Originally Posted by Jhevon View Post
    ok, so you used the product rule. but note that you have the wrong derivative for the inside.

    the derivative of 1 + 49.1e^{-0.5x} is -24.55e^{-0.5x}, there's your problem.

    remember, you have a constant 49.1 multiplying the e
    Ahh, of course, thanks a bunch mate!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simple derivative...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 23rd 2011, 02:40 PM
  2. simple derivative
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 27th 2009, 09:04 AM
  3. Simple Derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 7th 2009, 05:56 AM
  4. simple derivative of...
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 26th 2008, 05:16 PM
  5. Very very simple derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 22nd 2008, 04:07 AM

Search Tags


/mathhelpforum @mathhelpforum