Results 1 to 3 of 3

Math Help - Find f'(0) for f(x) = (e^x + e^-x)/2

  1. #1
    Banned
    Joined
    Oct 2009
    Posts
    56

    Find f'(0) for f(x) = (e^x + e^-x)/2

    Here's my answer...

    f'(x) = [(g'(x)) . (h(x)) - g(x) . h'(x)]/ (h(x))^2
    = [(e^x - e^-x) (2) - (e^x + e^-x) (0)]/4
    = 2(e^x - e^-x)/4 = (e^x - e^-x)/2

    f'(0) = [e^0 - e^0]/2 = 0



    Did I do that right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by satx View Post
    Here's my answer...

    f'(x) = [(g'(x)) . (h(x)) - g(x) . h'(x)]/ (h(x))^2
    = [(e^x - e^-x) (2) - (e^x + e^-x) (0)]/4
    = 2(e^x - e^-x)/4 = (e^x - e^-x)/2

    f'(0) = [e^0 - e^0]/2 = 0



    Did I do that right?
    i believe you did it right
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by satx View Post
    Here's my answer...

    f'(x) = [(g'(x)) . (h(x)) - g(x) . h'(x)]/ (h(x))^2
    = [(e^x - e^-x) (2) - (e^x + e^-x) (0)]/4
    = 2(e^x - e^-x)/4 = (e^x - e^-x)/2

    f'(0) = [e^0 - e^0]/2 = 0



    Did I do that right?
    Very inefficient and plenty of opportunity to make careless mistakes.

    Since f(x) = \frac{e^x}{2} + \frac{e^{-x}}{2}, why not just differentiate term-by-term?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: March 22nd 2011, 04:57 PM
  2. Replies: 2
    Last Post: July 5th 2010, 08:48 PM
  3. Replies: 1
    Last Post: February 17th 2010, 03:58 PM
  4. Replies: 0
    Last Post: June 16th 2009, 12:43 PM
  5. Replies: 2
    Last Post: April 6th 2009, 08:57 PM

Search Tags


/mathhelpforum @mathhelpforum