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I need to determine whether the following function is differentiable at $x_0 \ne 0$ according to the derivative definition:

$$ f(x) = x^\frac{1}{3} $$

So I started by looking at the definition but not sure how to proceed: $$ f'(x_0) =\lim_{x \to x_0} \frac{x^\frac{1}{3} - x_0^\frac{1}{3}}{x - x_0} $$

Thank you

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Do u know the formula of a^3 - b^3 ? Then use it .rest is easy.

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    I know it but not sure how to apply it here.. – 2017-01-20
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    @Noam, just factor $x - x_0$ = $(x^\frac13)^3 - ({x_0}^\frac13)^3$ as difference of cubes. – 2017-01-20
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    Put x^(1/3) =1 and x_0^(1/3)=b – 2017-01-20