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Math Help - Differentiation proof

  1. #1
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    Differentiation proof

    For a postive integers  n

    we have  ( x^n)' = nx^{n-1}

    for a fraction  \frac{p}{q}

    we also have  (x^{\frac{p}{q}})' = \frac{p}{q} x^{\frac{p}{q} -1}

    I can complete the above proofs but don't know to prove for irrational numbers index  n ? Is  (x^{n})' = n x^{n-1} ?

    Moreover , are  (x^{\pi} )' = \pi x^{\pi -1} and  (x^{e} )' = e x^{e -1} ?
    Thank you !
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  2. #2
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    Quote Originally Posted by simplependulum View Post
    For a postive integers  n

    we have  ( x^n)' = nx^{n-1}

    for a fraction  \frac{p}{q}

    we also have  (x^{\frac{p}{q}})' = \frac{p}{q} x^{\frac{p}{q} -1}

    I can complete the above proofs but don't know to prove for irrational numbers index  n ? Is  (x^{n})' = n x^{n-1} ?

    Moreover , are  (x^{\pi} )' = \pi x^{\pi -1} and  (x^{e} )' = e x^{e -1} ?
    Thank you !
    Proof of the Power Rule
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  3. #3
    ynj
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    Quote Originally Posted by simplependulum View Post
    For a postive integers  n

    we have  ( x^n)' = nx^{n-1}

    for a fraction  \frac{p}{q}

    we also have  (x^{\frac{p}{q}})' = \frac{p}{q} x^{\frac{p}{q} -1}

    I can complete the above proofs but don't know to prove for irrational numbers index  n ? Is  (x^{n})' = n x^{n-1} ?

    Moreover , are  (x^{\pi} )' = \pi x^{\pi -1} and  (x^{e} )' = e x^{e -1} ?
    Thank you !
    First we have: x^a,ax^{a-1}\in C[-\infty,+\infty](a is the indepedent variable).
    Let a be a irrational number. Let \{a_n\}be a rational sequence with \lim a_n=a.
    We get a sequence of functions \{f_n=x^{a_n}\}.
    Now we are to prove that \lim f'_n=(\lim f_n)', so that (x^a)'=\lim a_nx^{a_n-1}=ax^{a-1}.
    There is a theorem saying that:
    If 1: f_i\in C^{1}[c,d],
    2: \{f'_n\}uniformly converges at g,
    3: \{f_n\}converges at at least one point x_0,
    then \lim f'_n=(\lim f_n)'.
    we know that for every x, \lim x^{a_n}=x^{a},so 3 is true.
    1 is clearly true.
    Now we only have to show 2.Let g(x)=ax^{a-1}
    let b_n=\sup_{x\in [c,d]}\{|f'_n(x)-g(x)|\}, it suffices to show that \lim b_n=0
    Let e_n=a_n-a, then \lim e_n=0
    so a_nx^{a_{n-1}}-ax^{a-1}=e_nx^{a_{n-1}}+a(x^{a_{n-1}}-x^{a-1})
    We have x\in [c,d],|x^{a_{n-1}}|<M.
    so \forall\epsilon>0,\exists N,\forall n>N,|e_n|<\epsilon,|x^{a_{n-1}}-x^{a-1}|<\epsilon\forall x\in [c,d],the second inequality is true since x^{p-1}is uniformly continous on [c,d].
    So |a_nx^{a_{n-1}}-ax^{a-1}|<M\epsilon+|a|\epsilon=(M+|a|)\epsilon,we are done..
    So tough a proof...
    Last edited by ynj; September 3rd 2009 at 06:29 AM.
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  4. #4
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    Or, if you first prove that the derivative of ln(x) is 1/x, starting from y= x^r, take the ln of both sides: ln(y)= r ln(x).

    Now, differentiating both sides with respect to x, \frac{1}{y}\frac{dy}{dx}= \frac{r}{x}. Multiply both sides by y: \frac{dy}{dx}= \frac{r}{x}y= \frac{r}{x}x^r= r x^{r-1}.

    That's easier than ynj's proof but requires first proving that the derivative of ln(x) is 1/x as well as the chain rule.
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  5. #5
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