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Mar 11, 2024 •559 likes •686 views

Section 11.4 d i E The Product Rule and the Quotient Rule a l l u d Dr. Abdulla Eid b A College of Science . r D MATHS 104: Mathematics for Business II Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 1 / 12

  1. Section 11.4 d i E The Product Rule and the Quotient Rule a l l u d Dr. Abdulla Eid b A College of Science . r D MATHS 104: Mathematics for Business II Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 1 / 12

  2. Motivation Goal: We want to derive rules to find the derivative of product f ( x ) g ( x ) and quotient f ( x ) g ( x ) of two functions. d i E Example a l We want to find (in a general way) the derivative of the functions: l u d f ( x ) = ( 3 x + 1 )( 5 x + 2 ) . b A f ( x ) = xe x . f ( x ) = x 2 ln x . . r D 3 x + 1 f ( x ) = x 3 + 2 x + 1 . ln x f ( x ) = e x + 6 x − 3 . Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 2 / 12

  3. The Product Rule Theorem d ( f ( x ) g ( x )) ′ = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) . i E a l ( f ( x ) g ( x )) ′ = ( derivative of first ) ( second ) + ( first )( derivative of second ) l u d b A Before we prove this theorem, recall that the definition of the derivative is . r D f ( x + h ) − f ( x ) g ( x + h ) − g ( x ) f ′ ( x ) = lim and g ′ ( x ) = lim h h h → 0 h → 0 Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 3 / 12

  4. Proof: Let F ( x ) = f ( x ) g ( x ) . Then, F ( x + h ) − F ( x ) F ′ ( x ) = lim h h → 0 f ( x + h ) g ( x + h ) − f ( x ) g ( x ) d = lim i h E h → 0 a l l u We will use a “trick“ by adding and subtracting f ( x ) g ( x + h ) in the d b middle of the numerator. A . r D f ( x + h ) g ( x + h ) − f ( x ) g ( x + h ) + f ( x ) g ( x + h ) − f ( x ) g ( x ) F ′ ( x ) = lim h h → 0 [ f ( x + h ) − f ( x )] g ( x + h ) + f ( x )[ g ( x + h ) − g ( x )] = lim h h → 0 Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 4 / 12

  5. Continue... [ f ( x + h ) − f ( x )] g ( x + h ) + f ( x )[ g ( x + h ) − g ( x )] d = lim i h E h → 0 [ f ( x + h ) − f ( x )] g ( x + h ) + f ( x )[ g ( x + h ) − g ( x )] a = lim l l h h h → 0 u d [ f ( x + h ) − f ( x )] g ( x + h ) f ( x )[ g ( x + h ) − g ( x )] b = lim + lim h A h h → 0 h → 0 f ( x + h ) − f ( x ) g ( x + h ) − g ( x ) . = lim h → 0 g ( x + h ) + lim lim h → 0 f ( x ) lim r h D h h → 0 h → 0 = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 5 / 12

  6. Example Find the derivative of each of the following: 1 F ( x ) = ( x 2 + 5 x − 6 )( 6 x 2 − 5 x + 6 ) 2 F ( x ) = 2 ( √ x + 5 x − 3 )( 4 √ x − 4 √ x ) d i Solution: (1) E a F ′ ( x ) = ( derivative of first ) ( second ) + ( first )( derivative of second ) l l u = ( 2 x + 5 )( 6 x 2 − 5 x + 6 ) + ( x 2 + 5 x − 6 )( 12 x − 5 ) d b A . (2) r D F ′ ( x ) = ( derivative of first ) ( second ) + ( first )( derivative of second ) � 1 √ x − 4 √ x ) + ( √ x + 5 x − 3 )( 1 4 � − 3 = 2 ( 2 √ x + 5 )( 4 2 √ x ) 4 x 4 − Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 6 / 12

  7. Product rule for 3 functions ( fg ) ′ = f ′ g + fg ′ d ( fgh ) ′ = f ′ gh + fg ′ h + fgh ′ i E a l l u Example d b Find the derivative of each of the following: A 1 F ( x ) = ( x − 1 )( x − 2 )( x 2 − 4 ) . r D Solution: F ′ ( x ) = ( 1 )( x − 2 )( x 2 − 4 ) + ( x − 1 )( 1 )( x 2 − 4 ) + ( x − 1 )( x − 2 )( 2 x ) Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 7 / 12

  8. The Quotient Rule d Theorem i E � f ( x ) � ′ = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) a . l g ( x ) g ( x ) 2 l u d ( denominator ) derivative of numerator − ( numerator )( derivative of denominato b A ( denominator ) 2 . r D To prove this theorem, we will use the product rule. Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 8 / 12

  9. Proof: Let F ( x ) = f ( x ) g ( x ) . We want to find F ′ ( x ) . For that we apply the product rule to F ( x ) g ( x ) = f ( x ) ( derivative of first )( second ) + ( first )( derivative of second ) = f ′ ( x ) d i E F ′ ( x ) g ( x ) + F ( x ) g ′ ( x ) = f ′ ( x ) a F ′ ( x ) g ( x ) = f ′ ( x ) − F ( x ) g ′ ( x ) l l u F ′ ( x ) = f ′ ( x ) − F ( x ) g ′ ( x ) d b g ( x ) A f ′ ( x ) − f ( x ) g ( x ) g ′ ( x ) . F ′ ( x ) = r D g ( x ) F ′ ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) ( g ( x )) 2 Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 9 / 12

  10. Example (Old Exam Question) Find the derivative of each of the following: 2 1 F ( x ) = 5 x + 1 d 2 F ( x ) = 1 − x 1 − x 3 i E a Solution: (1) l l u F ′ ( x ) = ( denominator ) derivative of numerator − ( numerator )( derivative of denominato d b ( denominator ) 2 A = ( 5 x + 1 )( 0 ) − ( 2 )( 5 ) . r ( 5 x + 1 ) 2 D − 10 = ( 5 x + 1 ) 2 Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 10 / 12

  11. Continue... Recall we want to find the derivative of F ( x ) = 1 − x 1 − x 3 . d F ′ ( x ) = ( denominator )( derivative of numerator ) − ( numerator )( derivative of i E ( denominator ) 2 a = ( 1 − x 3 )( − 1 ) − ( 1 − x )( − 3 x 2 ) l l u ( 1 − x 3 ) 2 d = − 1 + x 3 + x 2 − 3 x 3 b A ( 1 − x 3 ) 2 . r = − 1 + x 2 − 2 x 3 D ( 1 − x 3 ) 2 Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 11 / 12

  12. Example (Old Exam Question) The revenue of selling q units per month is given by r ( q ) = 500 q q + 16 . Find the marginal revenue at q = 4. Solution: d Marginal revenue = dr i E dq a = ( denominator )( derivative of numerator ) − ( numerator )( derivative of denominato l l u ( denominator ) 2 d b = ( q + 16 )( 500 ) − ( 500 q )( 1 ) A ( q + 16 ) 2 . r 8000 D = ( q + 16 ) 2 At q = 4, we have 8000 dr = ( 4 + 16 ) 2 = 20 dq q = 4 Dr. Abdulla Eid (University of Bahrain) Product and Quotient Rules 12 / 12

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