# Differentiate the function y=(3x^2-2)/(3x^2+2)

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### 2 Answers

The function that has to differentiated is y = (3x^2 - 2)/(3x^2 + 2)

y = (3x^2 - 2)*(3x^2 + 2)^(-1)

y' = [(3x^2 - 2)*(3x^2 + 2)^(-1)]'

y' = (3x^2 - 2)'*(3x^2 + 2)^(-1) + (3x^2 - 2)*[(3x^2 + 2)^(-1)]'

y' = 6x*(3x^2 + 2)^(-1) + (3x^2 - 2)*[(-1)*(3x^2 + 2)^(-2)]'*6x

y' = 6x*(3x^2 + 2)^(-1) - (3x^2 - 2)*(3x^2 + 2)^(-2)*6x

y' = 6x*[(3x^2 + 2 - 3x^2 + 2]/(3x^2 + 2)^(2)

y' = 24x/(3x^2 + 2)^(2)

**The required derivative is 24x/(3x^2 + 2)^(2)**

We notice that we'll have to determine 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 = 3x^2-2 => u' = 6x

We'll put v =3x^2+2=> v' = 6x

We'll substitute u,v,u',v' in the formula above:

f'(x) = [6x*(3x^2+2) - (3x^2-2)*6x]/(3x^2+2)^2

We'll factorize by 6x:

f'(x) = 6x(3x^2+2 - 3x^2 + 2)/(3x^2+2)^2

We'll combine and eliminate like terms inside brackets:

f'(x) = 6x *(4)/(3x^2+2)^2

**f'(x) = 24x/(3x^2+2)^2**