Results 1 to 6 of 6

Math Help - Differential calculus of Functions

  1. #1
    Junior Member
    Joined
    Oct 2010
    Posts
    61

    Differential calculus of Functions

    i got 2 Functions Y(x)=lxl^3 f(x)=x-1

    1) Prove that Y can be Differentiable In point 0

    2) use the
    information from ur Proof In 1 to Prove that the Function l(x-1)l^3
    are
    Differentiable In point 1

    thanks for helping.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    \lim_{h\rightarrow 0}\frac{Y(0+h)-Y(0)}{h}=\lim_{h\rightarrow 0}\frac{|h|^3}{h}.

    Now compute the limit from the left and the limit from the right separately. They both give 0, showing that the derivative of Y at 0 exists and is equal to 0.

    For the second one use the derivative theorem about the composition of 2 functions.

    If you need clarification on the right and left hand limits let me know.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2010
    Posts
    61

    HI thnaks for answer

    but i need clarification on the right and left hand limits.

    U mean from left lim -0 is still give 0
    like lim 0+ is also give 0
    ?

    thnaks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,403
    Thanks
    1485
    Awards
    1
    Quote Originally Posted by ZOOZ View Post
    but i need clarification on the right and left hand limits. U mean from left lim -0 is still give 0
    like lim 0+ is also give 0
    h < 0\, \Rightarrow \,\frac{{\left| h \right|^3 }}<br />
{h} =  - h^2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    Yes. If h is positive, then |h| is just h. If h is negative, then |h| is -h.

    So if h is positive \frac{|h|^3}{h} = \frac{h^3}{h} = h^2. So the limit from the right is 0^2 = 0.

    If h is negative, \frac{|h|^3}{h} = \frac{(-h)^3}{h} = -h^2. So the limit from the left is -0^2 = 0.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Oct 2010
    Posts
    61
    OK thanks so much i will try alone to Solve the 2) Question.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Differential Calculus Problem on a family of cubic functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 22nd 2011, 06:01 AM
  2. Differential Calculus
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: November 19th 2009, 07:44 AM
  3. Differential calculus?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 30th 2008, 08:58 AM
  4. Differential calculus
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 29th 2008, 03:10 AM
  5. Differential Calculus
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 2nd 2007, 07:11 AM

Search Tags


/mathhelpforum @mathhelpforum