Results 1 to 9 of 9

Math Help - Not Understanding Proof for Derivative.

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    8

    Not Understanding Proof for Derivative.

    If y' = f'(x)
    Then y = f(x) in one integral.
    Now Let g(x) be any other integral in y' = f'(x) that is g'(x) = f'(x). Show that g(x) can differ from f(x) by at most a constant
    Then Let : w' = f'(x) - g'(x) = 0
    Then the equation of w as a function of x must be w = constant
    hence we see that w = f(x) - g(x) equals g(x) = f(x) + constant.
    Because g(x) is any integral other than f(x) all integrals are given by y = f(x) + c



    How did they get from w = constant to g(x) = f(x) + constant algebraically?
    second.... they said Now Let g(x) be any other integral in y' = f'(x) that is g'(x) = f'(x). But if they are two dif integrals why did they set g'(x) = f'(x)? Doesn't setting something equal to something mean they are equal? Is there something that I am misunderstanding?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Mike012 View Post
    If y' = f'(x)
    Then y = f(x) in one integral.
    Now Let g(x) be any other integral in y' = f'(x) that is g'(x) = f'(x). Show that g(x) can differ from f(x) by at most a constant
    Then Let : w' = f'(x) - g'(x) = 0
    Then the equation of w as a function of x must be w = constant
    hence we see that w = f(x) - g(x) equals g(x) = f(x) + constant.
    Because g(x) is any integral other than f(x) all integrals are given by y = f(x) + c



    How did they get from w = constant to g(x) = f(x) + constant algebraically?
    second.... they said Now Let g(x) be any other integral in y' = f'(x) that is g'(x) = f'(x). But if they are two dif integrals why did they set g'(x) = f'(x)? Doesn't setting something equal to something mean they are equal? Is there something that I am misunderstanding?
    g'(x) = f'(x).

    Integrate both sides wrt x:

    g(x) = f(x) + C ....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    it's a basic fact from the derivative: if g'(x)=0 for all x, then g(x)=k.

    so in your problem if f'(x)=g'(x) for all x, then f'(x)-g'(x)=(f(x)-g(x))'=0, that implies f(x)-g(x) is constant, hence, the result.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2010
    Posts
    8
    How could we assume though that g'(x) = f'(x) ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2010
    Posts
    8
    This proof is almost as if the person made this proof was like....
    If there is a function say x^3 + c OR x^3
    Then by differentiating , both functons would be 3x^2
    Now the question pops up... If we had a differentiated function f'(x) without knowing the original function f(x)...
    How could we know that the integration of f'(x) differs by only a constant....
    Well if we set
    x^3 + c = x^3
    then by diff..
    = 3x^2 + c' = 3x^2
    which equals
    c' = 0....
    Am I correct?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by mr fantastic View Post
    g'(x) = f'(x).

    Integrate both sides wrt x:

    g(x) = f(x) + C
    mr. f,

    this is what his textbook was proving!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by Mike012 View Post
    How could we assume though that g'(x) = f'(x) ?
    What they're showing, essentially, is that if g'(x) = f'(x), then f(x) - g(x) = constant.

    So if g'(x) = f'(x), then f'(x) - g'(x) = 0.

    It seems to me that they should have defined w and not w', i.e., w(x)=f(x) - g(x),

    So that w'(x) = f'(x) - g'(x) = 0 ... and on with the rest of their proof.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Oct 2010
    Posts
    18
    This problem has been troubled me recently but I have found the solution.
    Suppose that y' = g' where both y' and g' are functions of x.
    It is required to prove that the difference between y and g does not vary with x. This proof is important because it justifies taking the integral of both sides of any equation.

    If y' = g', for all x
    y' - g' = 0, for all x
    (y-g)' = 0, for all x

    The set of functions whose derivative is always 0 are known as the constant functions. Since (y-g)' = 0, y-g is a constant function. Let it be denoted by C.

    y-g = C

    Adding g to both sides,

    y = g + C where C is a function such that its derivative with respect to x is always 0.

    In this proof we assume that y and g are continuously differentiable on the interval in question.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Krizalid gave you the exact same proof at post #3!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Intuitive understanding of Frechet derivative?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 21st 2011, 10:36 AM
  2. [SOLVED] Understanding A Function using its derivative?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 7th 2011, 10:39 PM
  3. Help understanding proof
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 9th 2010, 10:58 AM
  4. Basic derivative, understanding a question.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 30th 2010, 02:01 AM
  5. Help Understanding a Proof
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 23rd 2009, 08:37 PM

Search Tags


/mathhelpforum @mathhelpforum