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Product Rule

Product Rule. In this section we develop a formula for the derivative of a product of two functions. It is tempting to guess, as Leibniz did three centuries ago, that the derivative of a product is the product of the derivatives. Function f is the product of functions g and h.

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Product Rule

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  1. Product Rule • In this section we develop a formula for the derivative of a product of two functions. • It is tempting to guess, as Leibniz did three centuries ago, that the derivative of a product is the product of the derivatives.

  2. Functionfis the product offunctions g andh f(x) = g(x)x h(x) It would be nice if the rule were f ′ (x) = g′ (x)x h′(x) But, it is NOT!!!

  3. EXAMPLE 1 If g(x) = 5x2and h(x) = 2x andf(x) = (g ·h)(x) = g(x)× h(x) then 2 10x3 f(x) = 5x2×2x = Not true!!!!!!!

  4. ProductRule f ′(x) =(g • h) ′ (x ) = h(x) • g ′(x) + g(x) • h′ (x) OR This is not just the product of two derivatives.

  5. Example 1 continued From the previous slide h(x) = 2x h '(x) = 2 g(x) = 5x2 g' (x) = 10x Try the Product Rule f ′(x) =(g • h) ′ (x ) = h(x)• g ′(x)+ g(x)• h′ (x) (g • h) ′ (x ) = 2x(10x) + 5x2(2) = 20x2 + 10x2= 30x2 Remember: (f • g)(x) = (5x2)(2x) = 10x3 so (f • g) ′(x ) = 30x2

  6. Example2 • to show that (g • h) ′ (x) ≠ g ′(x) • h ′(x) BUT

  7. Example 2 Continued b) to show that (g • h) ′ (x ) = h(x)• g ′ (x) + g(x)• h′ (x)

  8. Example 3 Differentiateh(x) = (5x3 + 2x)(x – 2 )by 2 methods. Two ways Multiply it out: Product rule: h(x) = (5x3 + 2x)(x – 2 ) h(x) = (5x3 + 2x)(x –2 ) h ′ (x) = (x –2)(15x2 + 2) + (-2x –3) (5x3 + 2x) h(x) = 5x + 2x –1 h ′ (x) = 15 + 2x –2 – 10 – 4x –2 h ′(x) = 5 – x– 2 h ′ (x) =5 – x –2

  9. Practice Question 1 Differentiateh(x) = –4x(3x– 8)by 2 methods. Two ways Multiply it out: Product rule: h(x) = –4x(3x– 8) h(x) = –4x(3x– 8) h′(x) = (3x – 8)(–4) + (–4x)(3) h(x) = –12x2+ 32x h ′ (x) = –12x + 32 – 12x h′(x) = –24x+ 32 h ′ (x) =–24x+ 32

  10. Practice Question 2 Differentiate h(x) = x– 2 (x3 – 3x + 6) by 2 methods. Two ways Multiply it out: Product rule: h(x) = x–2 (x3 – 3x + 6) h(x) = x–2 (x3 – 3x + 6) h′(x)= (x3 – 3x + 6)(–2x– 3 ) + (x–2)(3x 2 – 3) h(x) = x – 3x –1 + 6x–2 h ′ (x) = –2 + 6x– 2 – 12x– 3+ 3 – 3x– 2 h′(x) = 1 + 3x–2 – 12x–3 h′(x) = 1 + 3x–2 – 12x–3

  11. Practice Question 3 Differentiate f(x) = (2x – 1)(x2 + 1)by 2 methods. Two ways Multiply it out: Product rule: f(x) = (2x – 1)(x2 + 1) f(x) = (2x – 1)(x2 + 1) f(x) = 2x3 + 2x – x2 – 1 f ′(x) = (x2 + 1)(2) + (2x – 1)(2x) f(x) = 2x3 – x2 + 2x– 1 f ′(x)= 2x2 + 2 + 4x2 – 2x f ′ (x) = 6x2 – 2x + 2 f ′ (x) = 6x2 – 2x + 2

  12. Practice Question 4 f(x) = h(x) • g (x) = (x2 + 3)(2x3 + 5x) Two ways Product rule: Multiply it out: f ′(x) =h(x) • g ′(x) + g(x) • h′ (x) f(x) = (x2 + 3)(2x3 + 5x) f(x) = 2x5 + 5x3 + 6x3 + 15x f ′(x) = (2x3 + 5x)(2x) + (x2 + 3)(6x2 + 5) f(x) = 2x5 + 11x3 + 15x f ′(x) = 4x4 + 10x2 + 6x4 + 5x2 + 18x2+ 15 f ′(x) = 10x4 + 33x2 + 15 f ′(x) = 10x4 + 33x2 + 15

  13. Example 4 Differentiate by using the Product Rule

  14. Example 5 Differentiate by using the Product Rule

  15. Example 6 Simplify by Dividing Out a Common Factor.f (x) = (3x3 + 4)(1 – 2x3) Common Factor is 3x2 f ′(x) = (1 – 2x3)(9x2) + (3x3 + 4)(–6x2) f ′ (x) =3x2[3 (1 – 2x3) +(3x3 + 4)(–2)] f ′ (x) =3x2(3 – 6x3 – 6x3 – 8) f ′ (x) =3x2(– 12x3 – 5 ) f ′ (x) =–3x2(12x3 + 5 )

  16. Example 6 continuedWhy Not Just Expand???? f (x) = (3x3 + 4)(1 – 2x3) f ′(x) = (1 – 2x3)(9x2) + (3x3 + 4)(–6x2) f ′ (x) = 9x2 – 18x5 – 18x5 –24x2 f ′ (x) =–36x5– 15x2= –3x2 (12x3 + 5) This is probably easier BUT when we learn the Chain Rulethe common factor method becomes more useful. Just wanted you think about it for now!!!

  17. Practice Question 5 f(x) = (x2 – 5)(x2 – 3) Product Rule f ′(x) =(x 2 – 3)(2x) + (x2 – 5)(2x) Factor Expand f ′(x) = 2x(x 2 – 3 + x2 – 5) f ′(x)= 2x 3 – 6x + 2x 3 – 10x = 4x 3 – 16x = 2x(2x2– 8) = 4x (x2 – 4) = 4x (x 2 – 4) = 4x (x– 2)(x + 2) = 4x(x – 2)(x+ 2)

  18. Example 7 Find the slope of the tangent to the given curve at the point whose x-coordinate is given. y = x 4 (4x 3 + 2), x = –1 = (4x3 + 2)(4x3) + x4(12x2) You may substitute x = –1 right away = [4(–1)3+ 2)(4(–1)3] + (–1)4[12(–1)2] = 20 or simplify then substitute = 16x6 + 8x3 + 12x6 = 16(–1)6+ 8(–1) 3 + 12(–1)6= 20 Slope of the tangent at x = –1 is 20

  19. m=20

  20. Complete the remaining questions for homework

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