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The quotient rule for differentation E. Kim 1 Warm up Use the product rule 1 to find the derivative of h ( x ) = (5 x 3 + 9 x )( xe x + 7 x 2 ) 1 In words, the derivative of a product is the derivative of the first factor times the second, plus

  1. The quotient rule for differentation E. Kim 1

  2. Warm up Use the product rule 1 to find the derivative of h ( x ) = (5 x 3 + 9 x )( xe x + 7 x 2 ) 1 In words, the derivative of a product is the derivative of the first factor times the second, plus the first factor times the derivative of the second. 2

  3. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? 3

  4. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: 3

  5. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) 3

  6. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: 3

  7. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: k ′ ( x ) g ( x ) + k ( x ) g ′ ( x ) = f ′ ( x ) 3

  8. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: k ′ ( x ) g ( x ) + k ( x ) g ′ ( x ) = f ′ ( x ) ◮ Replace k ( x ) as above: 3

  9. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: k ′ ( x ) g ( x ) + k ( x ) g ′ ( x ) = f ′ ( x ) ◮ Replace k ( x ) as above: k ′ ( x ) g ( x ) + f ( x ) g ( x ) g ′ ( x ) = f ′ ( x ) 3

  10. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: k ′ ( x ) g ( x ) + k ( x ) g ′ ( x ) = f ′ ( x ) ◮ Replace k ( x ) as above: k ′ ( x ) g ( x ) + f ( x ) g ( x ) g ′ ( x ) = f ′ ( x ) ◮ Rewrite fraction: 3

  11. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: k ′ ( x ) g ( x ) + k ( x ) g ′ ( x ) = f ′ ( x ) ◮ Replace k ( x ) as above: k ′ ( x ) g ( x ) + f ( x ) g ( x ) g ′ ( x ) = f ′ ( x ) ◮ Rewrite fraction: k ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) = g ( x ) f ′ ( x ) g ( x ) g ( x ) 3

  12. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: k ′ ( x ) g ( x ) + k ( x ) g ′ ( x ) = f ′ ( x ) ◮ Replace k ( x ) as above: k ′ ( x ) g ( x ) + f ( x ) g ( x ) g ′ ( x ) = f ′ ( x ) ◮ Rewrite fraction: k ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) = g ( x ) f ′ ( x ) g ( x ) g ( x ) ◮ Subtract on both sides: 3

  13. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: k ′ ( x ) g ( x ) + k ( x ) g ′ ( x ) = f ′ ( x ) ◮ Replace k ( x ) as above: k ′ ( x ) g ( x ) + f ( x ) g ( x ) g ′ ( x ) = f ′ ( x ) ◮ Rewrite fraction: k ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) = g ( x ) f ′ ( x ) g ( x ) g ( x ) ◮ Subtract on both sides: k ′ ( x ) g ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) g ( x ) 3

  14. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: k ′ ( x ) g ( x ) + k ( x ) g ′ ( x ) = f ′ ( x ) ◮ Replace k ( x ) as above: k ′ ( x ) g ( x ) + f ( x ) g ( x ) g ′ ( x ) = f ′ ( x ) ◮ Rewrite fraction: k ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) = g ( x ) f ′ ( x ) g ( x ) g ( x ) ◮ Subtract on both sides: k ′ ( x ) g ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) g ( x ) ◮ Divide on both sides: 3

  15. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: k ′ ( x ) g ( x ) + k ( x ) g ′ ( x ) = f ′ ( x ) ◮ Replace k ( x ) as above: k ′ ( x ) g ( x ) + f ( x ) g ( x ) g ′ ( x ) = f ′ ( x ) ◮ Rewrite fraction: k ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) = g ( x ) f ′ ( x ) g ( x ) g ( x ) ◮ Subtract on both sides: k ′ ( x ) g ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) g ( x ) ◮ Divide on both sides: k ′ ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x )] 2 3

  16. The Quotient Rule Question If k ( x ) = f ( x ) g ( x ) , what is k ′ ( x ) ? ◮ Mult by g ( x ) on both sides: k ( x ) g ( x ) = f ( x ) ◮ Differentiate each side: k ′ ( x ) g ( x ) + k ( x ) g ′ ( x ) = f ′ ( x ) ◮ Replace k ( x ) as above: k ′ ( x ) g ( x ) + f ( x ) g ( x ) g ′ ( x ) = f ′ ( x ) ◮ Rewrite fraction: k ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) = g ( x ) f ′ ( x ) g ( x ) g ( x ) ◮ Subtract on both sides: k ′ ( x ) g ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) g ( x ) ◮ Divide on both sides: k ′ ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x )] 2 The Quotient Rule � ′ � f ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x )] 2 g ( x ) 3

  17. The Quotient Rule ... in Newton notation... � ′ � f ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x )] 2 g ( x ) ... in Leibnitz notation... = g ( x ) d dx [ f ( x )] − f ( x ) d dx [ g ( x )] � f ( x ) � d [ g ( x )] 2 dx g ( x ) In the upcoming examples, I’ve tried to label the steps which are just algebra with the text “alg”. 4

  18. Example [B] Exercise: If k ( x ) = x 3 − 1 x 8 + 1 , find k ′ ( x ) . 5

  19. Example [B] Exercise: If k ( x ) = x 3 − 1 x 8 + 1 , find k ′ ( x ) . Solution: f ( x ) = x 3 − 1 and g ( x ) = x 8 + 1 k ′ ( x ) = ( x 8 + 1)( x 3 − 1) ′ − ( x 3 − 1)( x 8 + 1) ′ ( x 8 + 1) 2 = ( x 8 + 1)(3 x 2 − 0) − ( x 3 − 1)(8 x 7 + 0) ( x 8 + 1) 2 = ( x 8 + 1)(3 x 2 ) − ( x 3 − 1)(8 x 7 ) alg ( x 8 + 1) 2 = (3 x 10 + 3 x 2 ) − (8 x 10 − 8 x 7 ) alg x 16 + 2 x 8 + 1 = − 5 x 10 + 3 x 2 + 8 x 7 alg x 16 + 2 x 8 + 1 5

  20. Example [B] Exercise: If z ( x ) = e − x , find dz dx . 6

  21. Example [B] Exercise: If z ( x ) = e − x , find dz dx . Solution: z ( x ) = 1 e x , so use f ( x ) = 1 and g ( x ) = e x . dx = e x d dx [1] − 1 d dx [ e x ] dz ( e x ) 2 = e x · 0 − 1 · e x ( e x ) 2 = − e x alg ( e x ) 2 = − 1 alg e x In the future... We’ll gain a NEW way to find the derivative of z ( x ) = e − x 6

  22. Two Mnemonics � ′ � f ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x )] 2 g ( x ) 7

  23. Two Mnemonics � ′ � f ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x )] 2 g ( x ) Hi over Lo � ′ � hi lo 7

  24. Two Mnemonics � ′ � f ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x )] 2 g ( x ) Hi over Lo � ′ � hi = lo d(hi) − hi d(lo) lo lo lo 7

  25. Two Mnemonics � ′ � f ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x )] 2 g ( x ) Hi over Lo � ′ � hi = lo d(hi) − hi d(lo) lo lo lo Hi over Ho � hi � ′ ho 7

  26. Two Mnemonics � ′ � f ( x ) = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x )] 2 g ( x ) Hi over Lo � ′ � hi = lo d(hi) − hi d(lo) lo lo lo Hi over Ho � hi � ′ = ho d(hi) − hi d(ho) ho ho ho 7

  27. Example [B] Exercise: If f ( x ) = e x + x 5 1 + 2 x 3 , find f ′ ( x ) . 8

  28. Example [B] Exercise: If f ( x ) = e x + x 5 1 + 2 x 3 , find f ′ ( x ) . Solution: f ′ ( x ) = (1 + 2 x 3 ) · ( e x + x 5 ) ′ − ( e x + x 5 ) · (1 + 2 x 3 ) ′ (1 + 2 x 3 ) 2 = (1 + 2 x 3 ) · ( e x + 5 x 4 ) − ( e x + x 5 ) · (0 + 6 x 2 ) (1 + 2 x 3 ) 2 = (1 + 2 x 3 ) · ( e x + 5 x 4 ) − ( e x + x 5 ) · (6 x 2 ) alg (1 + 2 x 3 ) 2 = ( e x + 5 x 4 + 2 x 3 e x + 10 x 7 ) − (6 x 2 e x + 6 x 7 ) alg 1 + 4 x 3 + 4 x 6 8

  29. Example, on your own... [B] Exercise: If y = ( x − 1)( x 2 − 2 x ) , compute dy dx . x 4 9

  30. Example, on your own... [B] Exercise: If y = ( x − 1)( x 2 − 2 x ) , compute dy dx . x 4 Solution: dx = [ x 4 ] · d dx [( x − 1)( x 2 − 2 x )] − [( x − 1)( x 2 − 2 x )] · d dx [ x 4 ] dy [ x 4 ] 2 = [ x 4 ][(1)( x 2 − 2 x ) + ( x − 1)(2 x − 2)] − [( x − 1)( x 2 − 2 x )][4 x 3 ] PR [ x 4 ] 2 = [ x 4 ][( x 2 − 2 x ) + (2 x 2 − 4 x + 2)] − [ x 3 − 3 x 2 + 2 x ][4 x 3 ] alg [ x 4 ] 2 = [ x 4 ][3 x 2 − 6 x + 2] − [ x 3 − 3 x 2 + 2 x ][4 x 3 ] alg [ x 4 ] 2 = [3 x 6 − 6 x 5 + 2 x 4 ] − [4 x 6 − 12 x 5 + 8 x 4 ] alg x 8 = − x 6 + 6 x 5 − 6 x 4 alg alg = − x − 2 + 6 x − 3 − 6 x − 4 x 8 9

  31. Revisit the same example Exercise: If y = ( x − 1)( x 2 − 2 x ) , compute dy dx . x 4 2 no calculus yet... 10

  32. Revisit the same example Exercise: If y = ( x − 1)( x 2 − 2 x ) , compute dy dx . x 4 Second Solution: ◮ First FOIL out numerator: y = x 3 − 3 x 2 + 2 x x 4 ◮ More algebra 2 : y = x − 1 − 3 x − 2 + 2 x − 3 . ◮ Use Sum/Difference Rule and Power Rule to take derivative: dy dx = ( − 1) x − 2 − 3 · ( − 2) x − 3 + 2 · ( − 3) x − 4 alg = − x − 2 + 6 x − 3 − 6 x − 4 which simplifies to Same example, two methods Avoiding the Quotient Rule was MUCH FASTER!!! 2 no calculus yet... 10

  33. 11

  34. Example [B] Exercise: If f ( x ) = x 3 + 2 x , find f ′ ( x ) . 4 12

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