# Differentiate the function f(x)=(3x+14)/(3x^2+7x)

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### 2 Answers

f(x)=(3x+14)/(3x^2+7x).

We use u(x)/v(x) = {(u'(x)v(x)-u(x)v'(x)}/(v(x))^2 to differentiate the given expression.

f'(x) = {(3x+4)'(3x^2+7x) -(3x+14)(3x^2+7x)'}/(3x^2+7x)62.

f(x) = {3(3x^2+7x)-(3x+14)(6x+7)}/(3x^2+7x)^2.

f'(x) = {9x^2+21x-18x^2-21x-84x-98}/(3x^2+7x)^2.

f'(x) = {-9x^2-84x-98}/(3x^2+7x)^2.

f'(x) = -(9x^2+84x+98)/(3x^2+7x)^2.

Since the given function is a quotient, we'll apply the quotient rule to find it's first derivative:

(u/v)' = (u'*v - u*v')/v^2 (*)

We'll put u = 3x+14 => u' = 3

We'll put v = 3x^2 + 7x => v' = 6x + 7

We'll substitute u,v,u',v' in the formula (*):

f'(x) = [3(3x^2 + 7x) - (3x+14)(6x + 7)]/(3x^2+7x)^2

We'll remove the brackets:

f'(x) = (9x^2 + 21x - 18x^2 - 21x - 84x - 98)/(3x^2+7x)^2

We'll combine and eliminate like terms:

**f'(x) = -(9x^2 + 84x + 98)/(3x^2+7x)^2**