Results 1 to 5 of 5
Like Tree2Thanks
  • 1 Post By frick
  • 1 Post By topsquark

Thread: calculate : f"'(x)

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    596
    Thanks
    3

    calculate : f"'(x)

    if:
    f(sinx)=(sinx)

    calculate : f"(x)

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    993
    Thanks
    501

    Re: calculate : f"'(x)

    The function $f:\mathbb{R}\rightarrow \mathbb{R}$ given by

    $f(x)=x^2$

    is twice differentiable and satisfies the condition

    $f(\sin (x))=(\sin x)^2$ for all $x \in \mathbb{R}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,477
    Thanks
    890
    Awards
    1

    Re: calculate : f"'(x)

    Quote Originally Posted by dhiab View Post
    if:
    f(sinx)=(sinx)

    calculate : f"(x)

    Use the chain rule:

    Let y = sin(x). Then
    $\displaystyle f(y) = y^2$

    and
    $\displaystyle \dfrac{df}{dx} = \dfrac{df}{dy} \dfrac{dy}{dx} = 2y \cdot cos(x) = 2~sin(x)~cos(x) = sin(2x)$

    Can you finish?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2019
    From
    Kansas
    Posts
    5
    Thanks
    6

    Re: calculate : f"'(x)

    Quote Originally Posted by topsquark View Post
    Use the chain rule:

    Let y = sin(x). Then
    $\displaystyle f(y) = y^2$
    No! That would be correct if $\displaystyle f(x)= sin^2(x)$ but the original post said $\displaystyle f(sin(x))= sin^2(x)$. As Idea said, that implies that $\displaystyle f(x)= x^2$ so $\displaystyle f'(x)= 2x$ and $\displaystyle f''(x)= 2$.

    and
    $\displaystyle \dfrac{df}{dx} = \dfrac{df}{dy} \dfrac{dy}{dx} = 2y \cdot cos(x) = 2~sin(x)~cos(x) = sin(2x)$

    Can you finish?

    -Dan
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,477
    Thanks
    890
    Awards
    1

    Re: calculate : f"'(x)

    Quote Originally Posted by frick View Post
    No! That would be correct if $\displaystyle f(x)= sin^2(x)$ but the original post said $\displaystyle f(sin(x))= sin^2(x)$. As Idea said, that implies that $\displaystyle f(x)= x^2$ so $\displaystyle f'(x)= 2x$ and $\displaystyle f''(x)= 2$.
    Yes, like I said $\displaystyle f(sin(x)) = sin^2(x)$. Yeah. That's what I wrote.

    Thanks for the catch!

    -Dan
    Thanks from romsek
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Apr 2nd 2012, 07:06 PM
  2. Replies: 1
    Last Post: Sep 16th 2011, 01:08 AM
  3. Replies: 2
    Last Post: Apr 24th 2011, 07:01 AM
  4. Replies: 1
    Last Post: Oct 25th 2010, 04:45 AM
  5. Replies: 1
    Last Post: Jun 4th 2010, 10:26 PM

/mathhelpforum @mathhelpforum