Is the function x^(1/3) differentiable everywhere? why?

Question

Is the function x^(1/3) differentiable everywhere? why?

Answer 1

The derivative is (1/3) x^-2/3 and this can not

be evaluated for x=0.  So it's not differentiable

in x=0.

Answer 2

The function is not differentiable at x = 0, because there is a vertical tangent line there.

The form of the derivative suggests this, in fact: (1/3)x^(-2/3) isn't defined when x = 0.

If you want to, you can look at the definition of the derivative and show directly that the limit does not exist:

lim h->0 [[f(0 + h) - f(0)] / h]

= lim h->0 (h^(1/3) / h)

= lim h->0 h^(-2/3)

= lim h->0 [1 / h^(2/3)]

= ∞.

So the limit does not exist.

So the function is not differentiable at 0.

Answer 3

This function is differentiable everywhere .Because It does not have any singular points.