Results 1 to 3 of 3

Math Help - reviewing for Finals and STUCK on this! It's not hard, just tricky

  1. #1
    Junior Member
    Joined
    Jul 2010
    Posts
    59

    reviewing for Finals and STUCK on this! It's not hard, just tricky

    Let F(x) = f(f(x)) and G(x)=(F(x))2 . You also know that f(9)=7, f(7)=3, f′(7)=5, f′(9)=7.


    Find F′(9)= find Find G′(9)=

    The answers are 35 and 210

    The answers are right, but I don't remember how I got this right back when I did it weeks ago? Any explanation will be great! thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by softballchick View Post
    Let F(x) = f(f(x)) and G(x)=(F(x))2 . You also know that f(9)=7, f(7)=3, f′(7)=5, f′(9)=7.


    Find F′(9)= find Find G′(9)=

    The answers are 35 and 210

    The answers are right, but I don't remember how I got this right back when I did it weeks ago? Any explanation will be great! thanks

    The chain rule applied:

    F'(9)=[f(f(9))]'=f'(9)f'(f(9))=7 f'(7)=7\cdot 5=35

    G'(9)=[F(9)^2]'=2F'(9)F(9)=2\cdot 35 f(f(9))=70f(7)=70\cdot 3=210

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by softballchick View Post
    Let F(x) = f(f(x)) and G(x)=(F(x))2 . You also know that f(9)=7, f(7)=3, f′(7)=5, f′(9)=7.


    Find F′(9)= find Find G′(9)=

    The answers are 35 and 210

    The answers are right, but I don't remember how I got this right back when I did it weeks ago? Any explanation will be great! thanks
    I'll do the first question leaving the second for you:

    To differentiate F(x)=f(f(x)) you have to use the chain-rule:

    F'(x)=f'(f(x)) \cdot f'(x)

    Now plug in x = 9:

    F'(9)=f'(f(9)) \cdot f'(9) According to the given values you'll get:

    F'(9)=f'(7) \cdot 7 which yields finally

    F'(9)= 5 \cdot 7 = 35

    ... and now it's your turn
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. I'm stuck with this hard probability question
    Posted in the Statistics Forum
    Replies: 1
    Last Post: December 11th 2009, 12:35 PM
  2. Replies: 4
    Last Post: October 7th 2009, 09:36 PM
  3. Integration by reduction: stuck hard
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 17th 2008, 12:58 AM
  4. Not a hard problem but stuck
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 17th 2008, 01:15 PM
  5. Stuck on a tricky integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 5th 2007, 04:24 PM

Search Tags


/mathhelpforum @mathhelpforum