Use a factoring of a difference of cubes: a^3-b^3=(a-b)(a^2+ab+b^2), to help you calculate the derivative of f(x)=x^1/3 directly from the...

`f(x)= x^(1/3)`

`==>
f'(x)= lim_(h->0) (f(x+h) - f(x))/ h `

`==>
f'(x)= lim_(h->0) ((x+h)^(1/3) - x^(1/3))/
h`

`==> (a^3 - b^3) =
(a-b)(a^2+ab+b^2)`

`==> (a-b) = (a^(1/3) - b^(1/3))
(a^(2/3) + (ab)^(1/3) + b^(2/3))`

`==> (a^(1/3) -
b^(1/3)) = (a-b)/ (a^(2/3) + (ab)^(1/3) +
b^(2/3))`

`==> (x+h)^(1/3) - x^(1/3) = ( x+h - x)/
((x+h)^(2/3) + ((x+h)x)^(1/3) + x^(2/3))`

`==>
lim_(h->0) f(x)= lim_(h->0) ((h/((x+h)^(2/3) + (x(x+h))^(1/3) +
x^(2/3)))/h)`

`==> lim_(h->0) f(x)=
lim_(h->0) (1/((x+h)^(2/3) + (x(x+h))^(1/3) +
x^(2/3)))`

`==> lim_(x->0) f(x) = 1/ (x^(2/3)
+ x^(2/3) + x^(2/3)) = 1/(3x^(2/3))`

`==>
lim_(h-> 0) f(x) = 1/3 x^-(2/3)`

`==> f'(x)=
1/3
x^-(2/3)`

``

``

``

``

``

``

``

``

``

``