Results 1 to 5 of 5

Math Help - Derivate

  1. #1
    Newbie
    Joined
    Dec 2010
    Posts
    15

    Derivate

    Hello everyone!

    I have to prove that the weak derivate of function h(x)=sinx on interval (0,2\pi) is function g(x)=cosx.
    Anybody know any hint how to do this?

    Thank you in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Have a go at

    \displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2010
    Posts
    15
    Ok so we get: \displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}=cosx. So the interval is not important here and weak derivate we prove at the same way like "normal" derivate?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by cantona View Post
    Ok so we get: \displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}=cosx. So the interval is not important here and weak derivate we prove at the same way like "normal" derivate?
    Use the fact that \sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A). This will allow you to simplify the left hand side and then cancel "h".
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by cantona View Post
    Ok so we get: \displaystyle \lim_{h\to 0 }\dfrac{\sin (x+h)- \sin x}{h}=cosx. So the interval is not important here and weak derivate we prove at the same way like "normal" derivate?

    If a function, h(x), is differentiable in the conventional sense, then it's weak derivative is equivalent to it's conventional derivative, g(x). That is to say any function, f(x), which which is equal to the its conventional derivative, g(x), except over some set of Lebesgue measure zero, is also a weak derivative of the function, h(x).

    I'm not sure the responses reflect that this refers to Lebesgue integration.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivate
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 16th 2011, 06:30 AM
  2. [SOLVED] How to derivate these?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 16th 2010, 01:56 PM
  3. How to derivate this one?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 9th 2010, 06:00 PM
  4. derivate
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 17th 2010, 03:04 PM
  5. Derivate.
    Posted in the Calculus Forum
    Replies: 10
    Last Post: July 6th 2006, 12:08 AM

Search Tags


/mathhelpforum @mathhelpforum