Episode 17: The Product Rule and the Quotient Rule

Review of Differentiation Rules

• Basic Building Block Rules
  • Definition of Derivative: ( f' = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} )
  • Derivative of a Constant: 0
  • Power Rule: Derivative of ( x^n ) is ( nx^{n-1} ) where ( n ) is any real number
• Constant Multiple Rule: ( (c \cdot f)' = c \cdot f' ) where ( c ) is a constant
• Sum and Difference Rules: ((f \pm g)' = f' \pm g')
• Chain Rule: ( (f(g(x)))' = f'(g(x)) \cdot g'(x) )

New Differentiation Rules

• Product Rule
  • To differentiate ( f(x) \times g(x) ), use: [ (f \cdot g)' = f , g' + g , f' ]
  • Remembered as: First times derivative of the second, plus second times derivative of the first.
• Quotient Rule
  • To differentiate ( \frac{f(x)}{g(x)} ), use: [ \left( \frac{f}{g} \right)' = \frac{g , f' - f , g'}{g^2} ]
  • Bottom times derivative of the top, minus top times derivative of the bottom, over bottom squared.

Example Applications

Product Rule Examples

• Function: ( f(x) = x^2 + 3x - 1 ) times ( 2x^2 - x - 5 )
  • Use product rule to find the derivative.
• Function: ( f(t) = (3t + 1) \times \sqrt{t^3 - 5t + 9} )
  • Utilize product rule and chain rule for the square root function.

Quotient Rule Examples

• Function: ( y = \frac{2x - 3}{4x + 1} )
  • Apply quotient rule, simplify the result.
• Function: ( f(x) = \frac{(2x - 1)^3}{(5x^2 + 7)^4} )
  • Use quotient rule and simplify using chain rule for embedded functions.
• Function: ( f(x) = \frac{x \cdot \sqrt{x^2 + 1}}{2x + 9} )
  • Demonstrates embedded product rule within a quotient rule problem.

Key Points

• The product rule and quotient rule allow the differentiation of more complex expressions involving products and quotients.
• Understanding and applying chain rule is essential for handling compositions within these problems.
• Simplification after applying the rules is often necessary to get the final answer.

Next Steps

• Explore how the newly learned rules enhance understanding of function behavior through their derivatives.