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Math Help - Differentiable Functions

  1. #1
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    Differentiable Functions

    I came across this proof in my notes and I could use some help getting started.

    Let n\in \mathbb{N}. Then, prove that x^n is differentiable at any point x and (x^n)'=nx^{n-1}.

    I know how to prove ones that are simpler such as f(x)=x but this one is a bit harder for me to figure out.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is...

    \displaystyle (x+h)^{n} = \sum_{k=0}^{n} \binom{n}{k}\ h^{k}\ x^{n-k} (1)

    ... so that...

    \displaystyle \lim_{h \rightarrow 0} \frac{(x+h)^{n} - x^{n} }{h} = \lim_{h \rightarrow 0} \sum_{k=1}^{n} \binom{n}{k}\ h^{k-1}\ x^{n-k} = n\ x^{n-1} (2)

    Kind regards

    \chi \sigma
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  3. #3
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    Thank you for your answer. However, this is much different notation then what my professor uses in class. I will show you the proof that he gave for a simpler problem and then maybe we can modify your solution to fit his notation...

    Let f(x)=x. Prove that f(x) is differentiable at every point x and f'(x)=1.

    Proof:
    \frac{f(x)-f(a)}{x-a}=\frac{x-a}{x-a}=1\rightarrow 1 as x\rightarrow a.
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  4. #4
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    Krizalid's Avatar
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    for the differentiability, the quotient \dfrac{x^n-a^n}{x-a} exists for x\to a since x-a always divides x^n-a^n, that's an algebra fact.

    as for computing its derivative, you were given a proof above.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Krizalid View Post
    for the differentiability, the quotient \dfrac{x^n-a^n}{x-a} exists for x\to a since x-a always divides x^n-a^n, that's an algebra fact.

    as for computing its derivative, you were given a proof above.
    To add to the two previous posters note that \displaystyle \frac{x^n-a^n}{x-a}=a^{n-1}\frac{\left(\frac{x}{a}\right)^n-1}{\frac{x}{a}-1}=a^{n-1}\sum_{k=0}^{n-1}\left(\frac{x}{a}\right)^{n-1}. As x\to a it's pretty clear this goes to na^{n-1}.
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