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Answer 1 · 2017-02-16T15:59:25

$d/dx( root(3)x ) = lim_(h->0) (root(3)(x+h)-root(3)(x))/h = 1/(3root(3)(x^2))$

Using the definition of derivative we have:

$d/dx( root(3)x ) = lim_(h->0) (root(3)(x+h)-root(3)(x))/h$

Now use the identity:

$(a^3-b^3) = (a-b)(a^2+ab+b^2)$

with $a= root(3)(x+h)$ and $b=root(3)(x)$ :

$( ( root(3)(x+h))^3 - (root(3)(x))^3) = (root(3)(x+h)-root(3)(x) )( root(3)((x+h)^2)+ root(3)(x(x+h)) + root(3)(x^2))$

$( x+h) -x = (root(3)(x+h)-root(3)(x) )( root(3)((x+h)^2)+ root(3)(x(x+h)) + root(3)(x^2))$

$h= (root(3)(x+h)-root(3)(x) )( root(3)((x+h)^2)+ root(3)(x(x+h)) + root(3)(x^2))$

So:

$(root(3)(x+h)-root(3)(x) )/h = 1/( root(3)((x+h)^2)+ root(3)(x(x+h)) + root(3)(x^2))$

and then:

$lim_(h->0) (root(3)(x+h)-root(3)(x))/h = lim_(h->0) 1/( root(3)((x+h)^2)+ root(3)(x(x+h)) + root(3)(x^2)) = 1/(3root(3)(x^2))$