07B-copyright Joe Kahlig Derivatives of Products and Quotients Page 1Derivatives of Products and QuotientsIn problems 1-8 find the derivative. You do no t have to sim-plify.1. y = (x7+ 3x4+ 5)(x8+ 7x + 1)2. y = (x9+1x5)(x−3− 2x−2)3. y = (x3+ 4x + 2)ex4. f(x) = (x3+ 5x2+ 1) log4x5. y =x4+ 7x2+ 8x5+ 5x3+ 66. y =3x+ 7xx4+ 7x3+ 57. f(x) =x4exx5+ 7x8. y =(x3+ 7x)(x2+ 1)x8+ 5x2+ 2Problems 9-12 re fer to the functions f and g that satisfy theproperties as shown in the table. Find the indicated quantity.x f(x) f′(x) g(x) g′(x)0 1 −3 3 51 2 6 7 112 −5 0 2 103 4 −1 −4 8Note: The table shows that f′(3) = −1, g(1) = 79. J′(0) if J (x) = f (x)g(x)10. H′(3) if H(x) = (x3− 7x + 3)g(x)11. K′(1) if K(x) =x3+ ln(x)f(x)12. M′(2) if M (x) =exf(x) + g(x)For problems 13-1 5, find the values of x where the tangentline is horizontal.13. y =−x + 2x2+ 1214. y =1x2− 6x + 1015. y =3x2+ 10x16. Find the va lue(s) of x where y =2x + 3x + 4has an instan-taneous rate of change of 5.17. Find the value(s) of x where y =−x + 3x − 5has an instan-taneous rate of change of 2.07B-copyright Joe Kahlig Derivatives of Products and Quotients Page 2Answers1. y′= (7x6+ 12x3)(x8+ 7x + 1) + (x7+ 3x4+ 5)(8x7+ 7)2. y′= (9x8−5x−6)(x−3−2x−2)+(x9+x−5)(−3x−4+4x−3)3. y′= (3x2+ 4)ex+ (x3+ 4x + 2)ex4. f′(x) = (3x2+ 10x) log4(x) + (x3+ 5x2+ 1)1x ln(4)5.y′=(x5+ 5x3+ 6)(4x3+ 14x) − (x4+ 7x2+ 8)(5x4+ 15x2)(x5+ 5x3+ 6)26.y′=(x4+ 7x3+ 5)(3xln(3) + 7) − (3x+ 7x)(4x3+ 21x2)(x4+ 7x3+ 5)27. f′(x) =(x5+ 7x)[4x3ex+ x4ex] − (x4ex)(5x4+ 7)(x5+ 7x)28. Step 1: foil the top of the fraction to make the derivativeeasier.y =x5+ 8x3+ 7xx8+ 5x2+ 2Step 2: take the derivativey′=(x8+ 5x2+ 2)(5x4+ 24x2+ 7) − (x5+ 8x3+ 7x)(8x7+ 10x)(x8+ 5x2+ 2)29. J′(x) = f′(x)g(x) + f (x)g′(x)J′(0) = f′(0)g(0) + f (0)g′(0)J′(0) = (−3)(3) + (1)(5)Answer: J′(0) = −410. H′(3) = −811. K′(x) =f(x)3x2+1x− (x3+ ln(x))f′(x)[f(x)]2K′(1) =f(1)3 +11− (1 + ln(1))f′(1)[f(1)]2Answer: K′(1) =24=1212. M′(2) =−13e2913. y′=(x2+ 12)(−1) − (−x + 2)(2x)(x2+ 12)2y′=x2− 4x − 12(x2+ 12)20 =x2− 4x − 12(x2+ 12)2Note: a fraction equal to zero means that the numeratormust be zero.0 = x2− 4x − 12Answer: x = −2, x = 614. y′=−2x + 6(x2− 6x + 10)2Answer: x = 315. y′=−6x − 30(x2+ 10x)2Answer: x = −516. y′=5(x + 4)25 =5(x + 4)2Now cross multiply, i.e. multiply both sides by (x + 4)25(x + 4)2= 5(x + 4)2= 1x + 4 = ±1Answer: x = −5, x = −317. x = 4, x =