# Find derivative of the fraction (3x^3+3)/(3x^3-3) .

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### 2 Answers

We have to find the derivative of y = (3x^3+3)/(3x^3-3)

We use the quotient rule which state that the value of [f(x) / g(x)]' = [f'(x)*g(x) - f(x)*g'(x)]/ [g(x)]^2

Substituting the functions we have:

[(3x^3+3)/(3x^3-3)]' = [(3x^3+3)'*(3x^3-3) - (3x^3+3)*(3x^3-3)']/ [(3x^3-3)]^2

=> [9x^2*(3x^3-3) - (3x^3+3)*9x^2]/ [(3x^3-3)]^2

=> [27x^5 - 27x^2 - 27x^5 - 27x^2] / [(3x^3-3)]^2

=> [- 27x^2 - 27x^2] / [(3x^3-3)]^2

=> [- 54x^2 ] / [(3x^3-3)]^2

=> [- 54x^2 ] / 9*[(x^3-1)]^2

=> [-6x^2 ] /[(x^3-1)]^2

**The required result is [-6x^2 ] /[(x^3-1)]^2**

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**