First, we'll factorize the numerator and denominator by
3:
(3x^3+3)/(3x^3-3) = 3(x^3 + 1)/3(x^3 -
1)
We'll simplify and we'll
get:
3(x^3 + 1)/3(x^3 - 1) = (x^3 + 1)/(x^3 -
1)
We notice that we'll have to find the derivative of a
fraction, so, we'll have to use the quotient rule.
(u/v)' =
(u'*v - u*v')/v^2
We'll put u = x^3 + 1 => u' =
3x^2
We'll put v = x^3 - 1 => v' =
3x^2
We'll substitute u,v,u',v' in the formula
above:
f'(x) = [3x^2*(x^3 - 1) - ( x^3 + 1)*3x^2]/(x^3 -
1)^2
We'll factorize by
3x^2:
f'(x) = 3x^2(x^3 - 1 - x^3 - 1)/(x^3 -
1)^2
We'll combine and eliminate like terms inside
brackets:
f'(x) = 3x^2 *(-2)/(x^3 -
1)^2
f'(x) = -6x^2/(x^3 -
1)^2
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