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Thread: Differentiation proof

  1. #1
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    Differentiation proof

    For a postive integers $\displaystyle n$

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

    for a fraction $\displaystyle \frac{p}{q} $

    we also have $\displaystyle (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 $\displaystyle n $ ? Is $\displaystyle (x^{n})' = n x^{n-1} $ ?

    Moreover , are $\displaystyle (x^{\pi} )' = \pi x^{\pi -1} $ and $\displaystyle (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 $\displaystyle n$

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

    for a fraction $\displaystyle \frac{p}{q} $

    we also have $\displaystyle (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 $\displaystyle n $ ? Is $\displaystyle (x^{n})' = n x^{n-1} $ ?

    Moreover , are $\displaystyle (x^{\pi} )' = \pi x^{\pi -1} $ and $\displaystyle (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 $\displaystyle n$

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

    for a fraction $\displaystyle \frac{p}{q} $

    we also have $\displaystyle (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 $\displaystyle n $ ? Is $\displaystyle (x^{n})' = n x^{n-1} $ ?

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

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

    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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