The derivative of F (x) = x ^ (4 / 5) - x ^ (9 / 5) does not exist when x =?

The derivative of F (x) = x ^ (4 / 5) - x ^ (9 / 5) does not exist when x =?

f'(x) = 4x^(-1/5)/5 - 9x^(4/5)/5
= 4/5x^(1/5) - 9x^(4/5) / 5
Because when x ^ 1 / 5, X cannot be equal to 0, otherwise 4 / 0 is meaningless!
So when x = 0, the derivative doesn't exist
This shows that the slope of the tangent line at this point may be n / 0, which is infinitely close to the vertical line

F (x) = {when x ≤ 1, (2 / 3) x ^ 3. When x > 1, x ^ 2} why do we say that the left derivative of F (z) at x = 1 exists and the right derivative does not exist

When x = 1, f (x) = 2 / 3
When X - > 1 +, f (x) - > 1
Discontinuous right derivative does not exist

Let y = f (x ^ 2 + b), where B is a constant and f has a second derivative

Yes, for the derivation of a function, the inner function (x ^ 2 + B is equivalent to the inner function, but it is not clearly marked) needs to be solved again.