09A-copyright Joe Kahlig The Chain Rule Page 1The Chain RuleProblems 1-5 refer to the functions f and g that satisfythe properties as shown in the table. Find the indicatedquantity.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 81. H′(0) if H(x) = f (g(x))2. J′(−1) if J(x) = f (x2)3. K′(2) if K(x) =x2+ g(x)34. J′(1) if J(x) = (x3+ 1)g(3x)5. H′(1) if H(x) = g(x2+ f(x))In problems 6-11, use loga rithm and exponential propertiesto simplify the function and then take the derivative.6. y = ln(x2+ 1)3(x + 5)57. y = ln(x3+ 1)2(x4+ 5)38. y = ln (3x + 5)5e3x2+5(2x + 1)3!9. y = eln(x3+7)+ eln(5+eln(x))10. y =ex2+4xe4x+511. y =ex4e3xex3−4For problems 12-14, find the values of x where the tange ntline is horizontal.12. y = x2(x − 3)413. y = (x + 1)2(x − 3)314. y = (x + 4)4(x − 3)3In problems 15-35, find each derivative. You do not haveto simplify.15. f (x) = (2x + 1)√x2+ 116. f (x) = e3x+ex17. f (x) = (x2+ 6x + 1)418. f (x) = 3x2+5x+119. f (x) = 1 + ln x + (ln x)2+ (ln x)320. f (x) = (x3+ 5x + 9)3221. f (x) =3qx3+1x322. f (x) = ln (ln (ln (x + 2)))23. f (x) = (ln x + xex+ 1)324. f (x) = [ln (x2+ 1)]4325. f (x) =ln(x) + 4e2x426. f (x) = (3x5− 1)4(x3+ 2)0.827. f (x) =(x4− 7x2)6+ 4x3 528. f (x) = e(x4+3x2+1)(2x3+ 7x)329. f (x) =e(x4+x2)+ ex4+ ex2(x4+ x2)30. f (x) = e(e2x)+ ln (ln (ln x))The derivatives is problems 31-35 are challenging andare messy. Do not simplify.31. f (x) = e√x4+3xln (x2+ 2x)32. f (x) =x23− 3x12+ 6x−45ex2+133. f (x) =ln xx2+ 13+ ex33x4+ 2x + 1234. f (x) ="7x4− x2x65+ (x2− 1)3(2x + x3)5#435. f (x) =sex2+e√x+ 1ln(x4+ 1) + 3209A-copyright Joe Kahlig The Chain Rule Page 2Answers1. H′(x) = f′(g(x)) ∗ g′(x)H′(0) = f′(g(0)) ∗ g′(0) = f′(3) ∗ 5 = −1 ∗ 5Answer: H′(0) = −52. J′(x) = 2xf′(x2)Answer: J′(−1) = − 123. K′(x) = 3x2+ g(x)2∗ (2x + g′(x))Answer: K′(2) = 15124. J′(x) = 3x2∗ g(3x) + (x3+ 1) ∗ 3g′(3x)Answer: J′(1) = 365. H′(x) = g′(x2+ f(x)) ∗ (2x + f′(x))Answer: H′(1) = 646. y = 3 ln(x2+ 1) + 5 ln(x + 5)y′= 3 ∗2xx2+ 1+ 5 ∗1x + 5=6xx2+ 1+5x + 57. y = 2 ln(x3+ 1) − 3 ln(x4+ 5)y′=6x2x3+ 1−12x3x4+ 58. y = 5 ln(3x + 5) + ln(e3x2+5) − 3 ln(2x + 1)y = 5 ln(3x + 5) + 3x2+ 5 − 3 ln(2x + 1)y′=153x + 5+ 6x −62x + 19. y = x3+ 7 + 5 + x = x3+ x + 12y′= 3x2+ 110. y = e(x2+4x)−(4x+5)= ex2−5y′= 2xex2−511. y′= (4x3− 3x2+ 3)ex4−x3+3x+412. y′= 2x(x − 3)4+ x2∗ 4(x − 3)3y′= 2x(x − 3)3)[(x − 3) + 2x]0 = 2x(x − 3)3(3x − 3)Answer: x = 0, x = 1, x = 313. y′= 2(x + 1) ∗ (x − 3)3+ (x + 1)2∗ 3(x − 3)2y′= (x + 1)(5x − 3)(x − 3)2Answer: x = −1, x = 3, x =3514. y′= 4(x + 4)3∗ (x − 3)3+ (x + 4)4∗ 3(x − 3)2y′= 7x(x − 3)2(x + 4)3Answer: x = 0, x = 3, x = −415. f′(x) = 2(x2+ 1)12+ (2x + 1)12(x2+ 1)−12(2x)16. f′(x) = (3 + ex)e3x+ex17. f′(x) = 4(x2+ 6x + 1)3(2x + 6)18. f′(x) = 3(x2+5x+1)(2x + 5)(ln 3)19. f′(x) =1x+2 ln xx+3(ln x)2x20. f′(x) = 1.5(x3+ 5x + 9)12(3x2+ 5)21. f′(x) =13(x3+ x−3)−23∗ (3x2− 3x−4)22. f′(x) =1ln(ln(x + 2))∗1ln(x + 2)∗1(x + 2)23. f′(x) = 3(ln x + xex+ 1)2∗1x+ ex+ xex24. f′(x) =43[ln (x2+ 1)]13∗2xx2+125. f′(x) = 4ln(x) + 4e2x3e2x1x− 2e2x(ln(x) + 4)(e2x)226. f′(x) = 4(3x5− 1)3(15x4) ∗ (x3+ 2)0.8+ 0.8(x3+ 2)−0.2(3x2) ∗ (3x5− 1)427. f′(x) = 5(x4− 7x2)6+ 4x3 4∗6(x4− 7x2)5(4x3− 14x) + 12x2 28. f′(x) = (4x3+ 6x)e(x4+3x2+1)∗ (2x3+ 7x)3+ 3(2x3+ 7x)2(6x2+ 7) ∗ e(x4+3x2+1)29. f′(x) =h(4x3+ 2x)e(x4+x2)+ (4x3)ex4+ (2x)ex2i(x4+ x2) +e(x4+x2)+ ex4+ ex2(4x3+ 2x)30. f′(x) = 2e2xe(e2x)+1(ln(ln(x)))1ln(x)1x31. f′(x) = 0.5(x4+ 3x)−0.5(4x3+ 3)e√x4+3x∗ ln(x2+ 2x) + e√x4+3x∗2x + 2x2+ 2x32. f′(x) =23x−13−32x−12+−245x−95e(x2+1)+x23− 3x12+ 6x−452xe(x2+1)33. f′(x) = 3ln(x)x2+ 12(x2+ 1)1x− ln(x) ∗ (2x)(x2+ 1)2+ 3x2ex3∗ (3x4+ 2x + 1)2+ ex3∗ (2)(3x4+ 2x + 1)(12x3+ 2)09A-copyright Joe Kahlig The Chain Rule Page 334. f′(x) = 4"7x4− x2x65+ (x2− 1)3(2x + x3)5#3"57x4− x2x64x6(28x3− 2x) − (7x4− x2)(6x5)(x6)2+3(x2− 1)2(2x) ∗ (2x + x3)5+ (x2− 1)3∗ (5)(2x + x3)4(2 + 3x2)35. f′(x) = 0.5 ex2+e√x+ 1ln (x4+ 1) + 32!−0.5∗"2xex2+12x−12e√x∗ln(x4+ 1) + 32+(e√x+ 1) ∗ (2)ln (x4+ 1) +