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Math Help - proving u has a derivative at x=0

  1. #1
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    Unhappy proving u has a derivative at x=0

    Hi all

    I ask for your opinion on my solution to the following problem i have been trying to solve for too long (3~ hours):

    "let w(x) be continuous for all x.
    we define u(x) = 2+x^2+xw(x)

    prove that u has a derivative at x=0."

    So i did it like this:

    u'(x) = 2x+w(x)+xw'(x)
    u'(0) = w(0)

    Since w is continuous for all x, w(0) exists, therefore u'(0) exists.


    I don't know why but it doesn't "feel" complete as a proof...it was "too easy" and something seems to be missing im not sure what.
    is it necessary to add that "the limit of u' as x approaches 0 = w(0) = u'(0)"?


    Thank you for your help.
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  2. #2
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    Re: proving u has a derivative at x=0

    Quote Originally Posted by ryu1 View Post
    "let w(x) be continuous for all x.
    we define u(x) = 2+x^2+xw(x)

    prove that u has a derivative at x=0."

    So i did it like this:
    u'(x) = 2x+w(x)+xw'(x)
    u'(0) = w(0)

    You have no way of knowing that w(x) has a derivative anywhere.

    But \frac{u(0+h)-u(0)}{h}=\frac{(2+h^2+hw(h))-(2)}{h}=\frac{h^2+hw(h)}{h}=~?\text{ as }h\to 0.
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  3. #3
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    Re: proving u has a derivative at x=0

    Nevermind, Plato said it better than I did.
    The Weierstrass function - Wikipedia, the free encyclopedia is a famous example of a function that is continuous everywhere but differentiable nowhere
    Last edited by Shakarri; April 20th 2013 at 08:09 AM.
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  4. #4
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    Re: proving u has a derivative at x=0

    Quote Originally Posted by Plato View Post
    You have no way of knowing that w(x) has a derivative anywhere.

    But \frac{u(0+h)-u(0)}{h}=\frac{(2+h^2+hw(h))-(2)}{h}=\frac{h^2+hw(h)}{h}=~?\text{ as }h\to 0.
    =h+w(h) = w(0)
    which is the same result(?)
    where do you go from here?
    why the way i suggested isn't right? (because it shows w'(x) which we can't know exists, inside u'(x)?)

    Thanks again.
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  5. #5
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    Re: proving u has a derivative at x=0

    Quote Originally Posted by ryu1 View Post
    =h+w(h) = w(0)
    which is the same result(?)
    where do you go from here?
    why the way i suggested isn't right? (because it shows w'(x) which we can't know exists, inside u'(x)?)
    Well, it is not the same result at all.
    Nowhere in that proof is it assumed that w' exists at all!

    It uses the definition of derivative: u'(0) = {\lim _{h \to 0}}\frac{{u(0 + h) - u(0)}}{h} = w(0).

    There is no more to do.
    Thanks from ryu1
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  6. #6
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    Re: proving u has a derivative at x=0

    u'(x) = 2x+w(x)+xw'(x)

    When you put x=0 into this expression for u'(x) you assumed that xw'(x) was zero but if w'(0) is undefined then 0*w'(0) is undefined.
    If the question is insisting that you prove it then write a little stating that it is impossible to prove. That knowing a function is continuous does not mean it is differentiable. Mention the Weierstrass function, your lecturer has probably heard of it.
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  7. #7
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    Re: proving u has a derivative at x=0

    Thank you Plato and Shakarri.

    @shakarri, the teacher did said that w(x) has no derivative and that we are supposed to prove u(x) has a derivative at 0 using w terms.
    (like the u'(0) = w(0))
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