Question #c8b78

1 Answer
Feb 16, 2017

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

Explanation:

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))