Quotient Rule – Differentiation of f(x) by g(x)

1. Concept Overview

When a function is written as a quotient of two differentiable functions:

[f(x)=\dfrac{u(x)}{v(x)}]

Quotient Rule Formula

[f'(x)][=\dfrac{v(x)\cdot u'(x) – u(x)\cdot v'(x)}{[v(x)]^{2}}]

Remember the order:
v u′ − u v′ ( NOT the opposite! )

Example 1

Differentiate: [f(x)=\dfrac{x^{2}+3x}{x+1}]

Step-by-Step Solution

1. Identify numerator and denominator:
   [u=(x^{2}+3x)], [v=(x+1)]
2. Derivatives:
   [u’=2x+3], [v’=1]
3. Apply quotient rule:
   [f'(x)=\dfrac{v u’ – u v’}{v^{2}}]
4. Substitute:
   [f'(x)][=\dfrac{(x+1)(2x+3) – (x^{2}+3x)(1)}{(x+1)^{2}}]
5. Expand and simplify numerator:
   [(x+1)(2x+3)][=2x^{2}+5x+3]
   Now subtract [x^{2}+3x]:
   [2x^{2}+5x+3 – (x^{2}+3x)][=x^{2}+2x+3]
6. Result:
   [f'(x)][=\dfrac{x^{2}+2x+3}{(x+1)^{2}}]

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{x^{2}+3x}{x+1}\big)][=\dfrac{x^{2}+2x+3}{(x+1)^{2}}]

Example 2

Differentiate: [f(x)=\dfrac{3x-1}{x^{2}}]

Step-by-Step Solution

1. [u=(3x-1)], [v=x^{2}]
2. [u’=3], [v’=2x]
3. Quotient rule:
   [f'(x)][=\dfrac{v u’ – u v’}{v^{2}}]
4. Substitute:
   [f'(x)][=\dfrac{x^{2}(3)-(3x-1)(2x)}{(x^{2})^{2}}]
5. Expand numerator:
   [x^{2}(3)=3x^{2}]
   [(3x-1)(2x)=6x^{2}-2x]
   Then:
   [3x^{2} – (6x^{2}-2x)][ = -3x^{2}+2x]
6. Result:
   [f'(x)=\dfrac{-3x^{2}+2x}{x^{4}}]
7. Simplify by dividing each term:
   [\dfrac{-3x^{2}}{x^{4}}][=-\dfrac{3}{x^{2}}]
   [\dfrac{2x}{x^{4}}][=\dfrac{2}{x^{3}}]

Conclusion:
[\dfrac{d}{dx}\big(\dfrac{3x-1}{x^{2}}\big)][=-\dfrac{3}{x^{2}}+\dfrac{2}{x^{3}}]