Is y = x2 / 3 (2 / 3 is power exponent, i.e. two thirds power of x) differentiable at x = 0 Y = x2 / 3 (2 / 3 is power exponent, i.e. two thirds power of x) Is it differentiable at x = 0 because the left derivative is negative infinity and the right derivative is positive infinity? Although the left and right derivatives are not equal, the tangents of them coincide at x = 0. I don't know whether it is differentiable On the first floor, I don't think your method is appropriate. For example: y = 0 (x = 0); y = x ^ 2 * sin (1 / x) (x / = 0), where x = 0 is y '= 2x * sin (1 / x) + cos (1 / x), X / = 0, then the point is not differentiable, But from the definition calculation, we can clearly get that the function is differentiable at x = 0, y '= LIM (x * sin (1 / x)) = 0, and y' tends to 0, but my question y 'can also be obtained from the definition as y' = LIM (x ^ (- 1 / 3)), and the left tends to negative infinity, the right tends to positive infinity, and the left is unequal

Non derivable
∵y′=（x^2/3）′=2/3x^-1/3
=2 / 3x ^ 1 / 3, where x ≠ 0
Non derivable