Given:

f(x) = y = (x^3)/3 - 2*x^(1/2)

Therefore:

Derivative of y = f'(x) = 3*(x^2)/3 - (2/2)*x^(-1/2)

= x^2 - 1/ (x^1/2)

Derivative of y for x = 3 is given by:

f'(3) = 3^2 - 1/(3^1/2)

= 9 - 0.5774 = 8.4226

To find the derivative of y = x^3/3 - 2sqrtx at x = 3.

To find dy/dx at x= 3 , we first find the derivative .Then find its value at x=3.

Let f(x) = y = x^3/3 - 2sqrtx.

f'(x) = dy/dx = {x^3/3 - 2x^(1/2)}'

dy/dx = (x^3/3)' - 2(x^1/2)'.

We use (cx^n)' =cn*x^(n-1) , where c is any constant but not zero.

dy/dx = 3*x^2/3 - 2(1/2)x^(1/2-1)

dy/dx = f'(x) = x^2-1/x^(1/2)

f'(3) = 3^2-1/3^(1/2)

f'(3) = 9 -(sqrt3)/3.

We'll calculate the expression of derivative of y.

We'll note f(x) = y.

f'(x) = (x^3/3 - 2sqrtx)'

f'(x) = 3x^2/3 - 2/2sqrtx

We'll simplify f'(x):

f'(x) = x^2 - 1/sqrtx

We'll substitute x by 3 and we'll get:

f'(3) = 3^2 - 1/sqrt3

f'(3) = 9 - 1/sqrt3** **

**Another method is to calculate the limit of the ratio:**

**lim [f(x) - f(3)/(x-3)],when x->3**

The value of the derivative of `y=x^3/3 - 2sqrtx` has to be determined at x=3.

If `f(x) = x^n, f'(x) = n*x^(n-1)`

y = `x^3/3 - 2sqrtx `

= `(1/3)*x^3 - 2*x^(1/2)`

y' = `(1/3)*3*x^2 - 2*(1/2)*x^(-1/2)`

= `x^2 - x^(-1/2)`

At x = 3, the value of `x^2 - x^(-1/2)` is `3^2 - 3^(-1/2) = 9 - 1/sqrt 3`

= `(9*sqrt 3 - 1)/sqrt 3`

The value of the derivative of `y=x^3/3 - 2sqrtx` at x=3 is `(9*sqrt 3 - 1)/sqrt 3`