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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! Please no fancy inverse function theorem, just the basic definition of the derivative...

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To show that f(x) is differentiable at x, you should prove that this limit exists:
lim[f(x+h)-f(x)]/h ; (h->0)
and then the result of this limit is f'(x). Therefore, we can narrow down the problem to finding this limit:

f'(x)=lim[(x+h)^(1/n)-x^(1/n)]/h ; (h->0)

Now, I show you how to find the limit, but before starting, observe that we will actually face 0/0 case:

1- l'Hopital's rule:

lim[(x+h)^(1/n)-x^(1/n)]/h= lim[(1/n)(x+h)^(1/n-1)]/1 ; (h->0)

You could easily find that this limit exists and it is actually:
f'(x)= (1/n)(x)^(1/n-1)=(1/n)(x)^([1-n]/n)

I personally like this method better because it is much easier to deal ...

Solution Summary

A function is proven to be differentiable using L'Hopital's Rule to find a limit.