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

    I know how to prove ones that are simpler such as $\displaystyle f(x)=x$ 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...

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

    ... so that...

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle f(x)=x$. Prove that $\displaystyle f(x)$ is differentiable at every point $\displaystyle x$ and $\displaystyle f'(x)=1$.

    Proof:
    $\displaystyle \frac{f(x)-f(a)}{x-a}=\frac{x-a}{x-a}=1\rightarrow 1$ as $\displaystyle x\rightarrow a$.
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  4. #4
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    Krizalid's Avatar
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    for the differentiability, the quotient $\displaystyle \dfrac{x^n-a^n}{x-a}$ exists for $\displaystyle x\to a$ since $\displaystyle x-a$ always divides $\displaystyle x^n-a^n,$ 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 $\displaystyle \dfrac{x^n-a^n}{x-a}$ exists for $\displaystyle x\to a$ since $\displaystyle x-a$ always divides $\displaystyle x^n-a^n,$ 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 $\displaystyle \displaystyle \frac{x^n-a^n}{x-a}=a^{n-1}\frac{\left(\frac{x}{a}\right)^n-1}{\frac{x}{a}-1}=a^{n-1}\sum_{k=0}^{n-1}\left(\frac{x}{a}\right)^{n-1}$. As $\displaystyle x\to a$ it's pretty clear this goes to $\displaystyle na^{n-1}$.
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