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    Prove that if f(x) = x^alpha, where alpha = 1/n for some n in N (the natural numbers), then y = f(x) is differentiable and f'(x) = alpha x^(alpha - 1).

    Progress I have made so far:

    I have managed to prove,

    (x^n)' = n x^(n - 1) for n in N and x in R

    both from the definition of differentiation involving the limit and the binomial theorem or equivalently using induction on n. Feel free to use this result although anything else should be made rigorous. It should be possible to prove this by the basic definition of the derivative. Thanks!

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    Solution Preview

    We first prove that (x^n)'=nx^(n-1) for all n, where n is a positive integer.

    Let f(x)=x^n and n is a positive integer. By the definition of f'(x) we have

    f'(x)=Limit{[f(x+h)-f(x)]/h, h->0} (1)

    where Limit{[f(x+h)-f(x)]/h, h->0} denotes the limit of ...

    Solution Summary

    Limits are used to prove differentiability.