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Math Help - Power Rule for derivatives

  1. #1
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    Power Rule for derivatives

    Let f(x)=x^n and prove that its derivative is given by f'(x)=n\cdot x^{n-1},\,\forall\,n\in\mathbb R.
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    There's a problem with that solution 'cause binomial theorem works with natural numbers, hence, when computing the limit, you assume that n is natural since n is a real number.
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    Yes, I know, but in that case we state that such derivative only works for natural numbers.
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    Hello,
    Quote Originally Posted by Krizalid View Post
    Yes, I know, but in that case we state that such derivative only works for natural numbers.
    Generalized Newton's binomial formula :

    (x+y)^n=\sum_{k=0}^\infty {n \choose k} x^k y^{n-k}

    where {n \choose k}=\frac{r(r-1)\dots(r-k+1)}{k!} and \forall n \in \mathbb{R}

    _______________________________________
    A way of doing it is to remember another formula for the derivative :
    (assuming it is differentiable so at least x_0 \neq 0)

    f'(x_0)=\lim_{x \to x_0} ~ \frac{x^n-x_0^n}{x-x_0}

    \frac{x^n-x_0^n}{x-x_0}=\frac{x_0^n}{x_0} \times \frac{\left(\frac{x}{x_0}\right)^n-1}{\frac{x}{x_0}-1}=x_0^{n-1} \times \sum_{k=0}^{n-1} \left(\frac{x}{x_0}\right)^k

    \lim_{x \to x_0} ~ x_0^{n-1} \times \sum_{k=0}^{n-1} \left(\frac{x}{x_0}\right)^k=x_0^{n-1} \times \sum_{k=0}^{n-1} 1^k=x_0^{n-1} \times n \quad \square
    Last edited by Moo; October 17th 2008 at 01:56 PM. Reason: inverted the binomial coeff again...
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    This is my approach:

    \begin{aligned}<br />
   \frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}&=\frac{e^{n\ln (x+\Delta x)}-e^{n\ln x}}{\Delta x} \\ <br />
 & =x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{\Delta x} \\ <br />
 & =x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}\cdot \frac{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}{\Delta x},<br />
\end{aligned}

    hence,

    \frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}=x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}\cdot \frac{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}{\dfrac{\Delta x}{x}}\cdot \frac{1}{x}\to nx^{n-1} as \Delta x\to0.\quad\blacksquare
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  6. #6
    Moo
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    Quote Originally Posted by Krizalid View Post
    This is my approach:

    \begin{aligned}<br />
   \frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}&=\frac{e^{n\ln (x+\Delta x)}-e^{n\ln x}}{\Delta x} \\ <br />
 & =x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{\Delta x} \\ <br />
 & =x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}\cdot \frac{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}{\Delta x},<br />
\end{aligned}

    hence,

    \frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}=x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}\cdot \frac{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}{\dfrac{\Delta x}{x}}\cdot \frac{1}{x}\to nx^{n-1} as \Delta x\to0.\quad\blacksquare
    Pardon me for that, but I do not find it elegant

    First of all, why don't you directly factor by x^n in the beginning instead of factorizing with exponentials ?

    Also, what if x+\Delta x < 0 ?

    Plus, you ought to know the limits \lim_{x \to 0} ~ \frac{\ln(1+x)}{x}=1 and \lim_{x \to 0} ~ \frac{e^x-1}{x}=1, which are proved using the definitions of derivative numbers (if not "advanced" calculus...)

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  7. #7
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    Quote Originally Posted by Moo View Post

    Pardon me for that, but I do not find it elegant
    Who said it was elegant?

    Quote Originally Posted by Moo View Post

    First of all, why don't you directly factor by x^n in the beginning instead of factorizing with exponentials ?
    To use known limits.

    Quote Originally Posted by Moo View Post

    Also, what if x+\Delta x < 0 ?
    Why should I consider that case?

    Quote Originally Posted by Moo View Post

    Plus, you ought to know the limits \lim_{x \to 0} ~ \frac{\ln(1+x)}{x}=1 and \lim_{x \to 0} ~ \frac{e^x-1}{x}=1, which are proved using the definitions of derivative numbers (if not "advanced" calculus...)
    By just knowing that \lim_{x\to\infty}\left(1+\frac1x\right)^x=e those limits are easy to prove.
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  8. #8
    Moo
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    Quote Originally Posted by Krizalid View Post
    Who said it was elegant?
    Just giving my point.

    To use known limits.
    You factorize later, after changing into exponentials, while it would have been exactly the same if you did it from the beginning.

    Why should I consider that case?
    \ln(x+\Delta x) when x + \Delta x< 0 ?


    By just knowing that \lim_{x\to\infty}\left(1+\frac1x\right)^x=e those limits are easy to prove.
    What's the interest of asking to prove a simple formula if it has to use nested limits ?

    Well, I'm no one to tell you this, so just ignore it if you want.
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    Krazalid is correct to say limit definition is no good if n is not an integer.

    And generalized binomial theorem is faulty since you implicity use the power rule in its derivation.

    The standard way is to notice that by definition x^r = \exp(r\ln x) for x>0.
    Here \exp is the inverse function for \ln t = \smallint_1^t \tfrac{du}{u}.

    If we are familar with the basic rules for \ln and \exp just apply the chain rule and get your answer.
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