# `f(x)=x^(2/3)`

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You need to find the derivative of the function `f(x)=x^(2/3)` using the first principle, such that:

`f'(x) = lim_(h-gt0) (f(x+h) - f(x))/h`

`f'(x) = lim_(h-gt0) (root(3)((x+h)^(2)) - root(3)(x^2))/h`

You need to remember that difference of cubes yields a product:

`x^3 - y^3 = (x-y)(x^2 +xy + y^2)`

Notice that you need to multiply `(root(3)(x^2+2hx + h^2) - root(3)(x^2))` by the conjugate `(root(3)((x+h)^(4))+ root(3)((x+h)^2(x^2)) +root(3)(x^4))` , hence:

`f'(x) = lim_(h-gt0) ((x+h)^(2) - (x^2))/(h(root(3)((x+h)^(4))+ root(3)((x+h)^2(x^2)) +root(3)(x^4)))`

Expanding the square `(x+h)^2` yields:

`f'(x) = lim_(h-gt0) (x^2+ 2xh + h^2 - x^2)/(h(root(3)((x+h)^(4))+ root(3)((x+h)^2(x^2)) +root(3)(x^4)))`

Reducing like terms yields:

`f'(x) = lim_(h-gt0) (2xh + h^2)/(h(root(3)((x+h)^(4))+ root(3)((x+h)^2(x^2)) +root(3)(x^4)))`

Factoring out h yields:

`f'(x) = lim_(h-gt0) (h(2x + h))/(h(root(3)((x+h)^(4))+ root(3)((x+h)^2(x^2)) +root(3)(x^4)))`

Reducing by h yields:

`f'(x) = lim_(h-gt0) (2x + h)/(root(3)((x+h)^(4))+ root(3)((x+h)^2(x^2)) +root(3)(x^4))`

Substituting 0 for h yields:

`f'(x) = (2x)/(3root(3)(x^4)) =gt f'(x) = (2x)/(3xroot(3)(x))`

Reducing by x yields:

`f'(x) = 2/(3root(3)(x))`

Hence, using the first principle of derivatives yields what you may get using the power property of derivatives: `f'(x) = (2/3)*x^(2/3 - 1) =gt f'(x) = 2/(3*x^(1/3)).`

**Hence, using the first principle of derivatives for `f(x) = x^(2/3)` yields that `f'(x) =2/(3root(3)(x))` .**