Results 1 to 8 of 8

Math Help - Deriviative of function #2

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    425

    Deriviative of function #2

    i)Find h'(2) given that h(x)=f(g(x)), f(u)=u^2-1, g(2)=3, g'(2)=-1

    ii) let y=f(x^2+3x-5). find dy/dx when x=1, given that f'(-1)=2
    Last edited by mr fantastic; August 12th 2009 at 04:02 PM. Reason: Changed post title
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19
    h(x)=f(g(x))

    f(u)=u^2-1

    f(g(x))=(g(x))^2-1

    Just use the chain rule:

    h'(x) = 2g(x)g'(x)

    You can now just substitute to get h'(2)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
     y(x) = f(x^{2}+3x-5)

     y'(x) = f'(x^{2}+3x-5) \frac{d}{dx} (x^{2}+3x-5) = f'(x^{2}+3x-5)(2x+3)

     y'(1) = f'(-1)(5) = 10
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19
    y=f(g(x)

    Let g(x)=x^2+3x-5

    \frac{dy}{dx}=f'((g(x))g'(x)

    =f'(g(x))[2x+3]

    When x=1, g(1)=-1, so we have \frac{dy}{dx}=5f'(-1)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    Quote Originally Posted by adkinsjr View Post
    y=f(g(x)

    Let g(x)=x^2+3x-5

    \frac{dy}{dx}=f'((g(x))g'(x)

    =f'(g(x))[2x+3]

    When x=1, g(1)=-1, so we have \frac{dy}{dx}=5f'(-1)
    where did the 5 come from in the last line
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19
    Quote Originally Posted by skeske1234 View Post
    where did the 5 come from in the last line
    The 5 is the derivative g'(x)=2x+3 at x=1
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    Quote Originally Posted by adkinsjr View Post
    h(x)=f(g(x))

    f(u)=u^2-1

    f(g(x))=(g(x))^2-1

    Just use the chain rule:

    h'(x) = 2g(x)g'(x)

    You can now just substitute to get h'(2)

    not sure what you mean.... i sorry don't follow that.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19
    Quote Originally Posted by skeske1234 View Post
    not sure what you mean.... i sorry don't follow that.
    We have the function f(u)=u^2-1 if we set u=g(x) we get the function h(x)=f(g(x))=(g(x))^2-1 (notice how I substituted u=g(x)). We use the chain rule to differentiate:

    h'(x)=f'(g(x))g'(x).

    =2g(x)g'(x).

    Do you understand how I got h'(x)=2g(x)g'(x)?

    If so, then we have

    h'(2)=2g(2)g'(2)

    h'(2)=2(3)(-1)=-6
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Deriviative of function
    Posted in the Calculus Forum
    Replies: 7
    Last Post: August 14th 2009, 01:23 PM
  2. Deriviative of function # 8
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 14th 2009, 01:01 PM
  3. Deriviative of function #1
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 8th 2009, 12:32 PM
  4. Deriviative of function #4
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 4th 2009, 11:45 AM
  5. Deriviative of function #5
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: August 4th 2009, 11:18 AM

Search Tags


/mathhelpforum @mathhelpforum