Решения к Сборнику заданий по высшей математике Кузнецова Л.А. – 2. Дифференцирование. Зад.5

Задача 5. Найти производную. 5.1. (9×2+8x-1)(x+1)1/2 – (3×3+4×2-x-2) y’=2/15* ___________________2(1+x)1/2 = 1+x = 2/15* (2x+2)(9×2+8x-1)-3×3-4×2+x+2 = 2(x+1)3/2 =2/15* 18×3+16×2-2x+18×2+16x-2-3×3-4×2+x+2 = 2(x+1)3/2 = 2/15* 15×3+30×2+15x = 2(x+1)3/2 = x(x+1)2 = x(x+1)1/2 (x+1)3/2 5.2. 3×3*4x(x2+1)1/2+x(2×2-1) -9×2(2×2-1)(x2+1)1/2 y’= (x2+1)1/2 = 9×6 = 12×4(x2+1)+3×4(2×2-1)-9×2(2×2-1)(1+x2) = 9×6(x2+1)1/2 = 12×4+12×6+6×6-3×4-18×4-18×6+9×2+9×4 = 9×6(x2+1)1/2 = 9×2 = 1 . 9×6(x2+1)1/2 x4(x2+1)1/2 5.3. y’= (4×3-16x)(x2-4)-(x4-8×2)2x = 4×5-16×3-16×3+64x-2×5+16×3 = 2(x2-4)2 2(x2-4)2 =2×5-16×3+64x =x(x2-4)2+16x = x+ 16×2 . 2(x2-4)2 (x2-4)2 (x2-4)2 5.4. (4x-1)√(2+4x) – 2(2×2-x-1) y’= √(2+4x) = (4x-1)(2+4x)-4×2+x+1 = 3(2+4x) 3(2+4x)√(2+4x) = 12×2+5x-1 . 3(2+4x)√(2+4x) 5. 5. 8×19√(1+x8)+ 4×19(1+x8) – 12×11(1+x8)3/2 y’= √(1+x8) = 12×24 = 12×19(1+x8)-12×11(1+x8)2 = 12×24√(1+x8) = x11(x16-2×8+1) = (x8-1)2 . x24√(1+x8) x13√(1+x8) 5.6. 2x√(1-3×4) + 6×5 ­ y’= √(1-3×4) = 2x(1-3×4)+6×5 = x . 2(1-3×4) 2(1-3×4)√(1-3×4) √(1-3×4)3 5.7. y= (2x(4+x2)√(4+x2)+3/2√(4+x2)*2x)x5-(x2-6)(4+x2)√(4+x2)*5×4 = 120×10 = √(4+x2)(8×6+2×8+3×6-20×6-5×8+30×6+120×4) = 120×10 = √(4+x2)(7×2-x4+40) 40×6 5.8. y= 3/2√(x2-8)*2×4-(x2-8)√(x2-8)*18×2 = 6×6 √(x2-8)(x4-6×4+48×2) = √(x2-8)(48-5×2) 3×6 3×4 5.9. 9×3(2+x3)2/3-(4+3×3)((2+x3)2/3+2/3* 3×3 ) y’= (2+x3)1/3 = x2(2+x3)4/3 = 9×3(2+x3)-(4+3×3)(2+3×3) = 8 . x2(2+x3)5/3 x2(2+x3)5/3 5.10. y’= √(x)*(2(1+x3/4)*3/4×5/4-(1+x3/4)2*3/2*√(x)) = 3(1+x3/4)2/3*x6/4 = √(x)(x3/2-1) 2x(1+x3/2)2/3 5.11. (6×5+3×2)√(1-x3) + 3×2(x6+x3-2) y’ = 2√(1-x3) = 1-x3 =(2-2×3)(6×5+3×2)+3×8+3×5-6×2 = (9×5-9×8) = 9×5 . 2(1-x3)3/2 2(1-x3)3/2 2√(1-x3) 5.12. 2×4√(4+x2)+ x4(x2-2) -3×2(x2-2)√(4+x2) y’= √(4+x2) = 24×6 = 2×4(4+x2)+x4(x2-2)-3×2(x2-2)(4+x2) = 1 24×6 x4 5.13. 2x√(1+2×2)- 2x(1+x2) y’= √(1+2×2) = x(1+2×2)-x(1+x2) = x3 . 2(1+2×2) (1+2×2)3/2 (1+2×2)3/2 5.14. y’= ((3x+2)/(2√(x-1))+3√(x-1))x2-2x√(3x+2) = 4×4 = x2(3x+2)+6×2(x-1)-4x(x-1)(3x+2) = 9×3-12×2+8x = 9×2-12x+8 4×2√(x-1) 4×2√(x-1) 4x√(x-1) 5.15. y’= 3/2*√(1+x2)*2×4-3×2(1+x2)3/2 = √(1+x2)*(x4-x2-x4) = -√(1+x2) 3×6 x6 x4 5.16. (6×5+24×2)√(8-x3)+3×2(x6+8×3-128) y’= 2√(8-x3) = 8-x3 = (16-2×3)(6×5+24×2)+3×2(x6+8×3-128) = 72×5-9×8 = 9×5 2(8-x3)3/2 2(8-x3)3/2 2√(8-x3) 5.17. x2(x-2)+x2√(2x+3)-(2×2-4x)√(2x+3) y’= √(2x+3) = x4 = x2(x-2+2x+3)-(2×2-4x)(2x+3) = 3×2-x3+12x = 3x-x2+12 x4√(2x+3) x4√(2x+3) x3√(2x+3) 5.18. y’=-2×5√(x3+1/x)+(1-x2)*1/5*(x3+1/x)4/5*(3×2-1/x2)=1/5*(x3+1/x)4/5(3×2-1/x2-3×4+1)-2x(x3+1/x)1/5 5.19. 4×4√(x2-3)+x4(2×2+3) – 3×2(2×2+3)√(x2-3) y’ = √(x2-3) = 9×6 = 4×4(x2-3)+x4(2×2+3)-3×2(2×2+3)(x2-3) = 27×2 = 3 . 9×6√(x2-3) 9×6√(x2-3) x4√(x2-3) 5.20. y’= (x2+5)3/2-3/2*(x-1)√(x2+5)*2x = √(x2+5)(5+3x-2×2) (x2+5)3 (x2+5)3 5.21. 2×2√(x2-x)+(2x-1)(2x+1)x2-2x(2x+1)√(x2-x) y’= √(x2-x) = x4 = x2(2×2-2x+4×2-1)-(4×2+2x)(x2-x) = 2×2+1 x4 x2 5.22. _ 1+√x _ 1-√x y’ = √((1+√x)/(1-√x))* 2√x 2√x = (1+√x)2 = -2√((1+√x)/(1-√x)) = -1 . 2√x(1+√x)2 √(x(1-x))(1+√x) 5.23. √(x2+4x+5) – x(x+2) y’ = √(x2+4x+5) = – 2×2-6x-5 . (x+2)2(x2+4x+5) (x+2)2(x2+4x+5)3/2 5.24. 2x+1 -3(x2+x+1)1/3 y’ = (x2+x+1)2/3 = -3×2-x-2 . (x+1)2 (x+1)2(x2+x+1)2/3 5.25. y’= 3√((x-1)4/(x+1)2)*(x-1)2-2(x-1)(x+1) = -3√((x-1)4/(x+1)2)*x2+2x-3 = (x-1)4 (x-1)4 = 3-x2-2x (x2-1)2/3(x-1)2 5.26. √(x2+2x+7)-(x+1)(x-1) y’ = √(x2+2x+7) = x2+2x+7-x2-8x-7 = -x . 6(x2+2x+7) 6(x2+2x+7)3/2 (x2+2x+7)3/2 5.27. y’ = (x2+x+1)(√(x+1)+x/(2√(x+1)))-(2×2+x)√(x+1) = (x2+x+1)2 = (3x+2)(x2+x+1)-(4×2+2x)(x+1) = -x3-x2+3x+2 2(x2+x+1)√(x+1) 2(x2+x+1)√(x+1) 5.28. y’ = 2x√(1-x4)+2x(x2+2)/√(1-x4) = 3x-x5+x3 2-2×4 (1-x4)3/2 5.29. y’ = (√(2x-1)+(x+3)/√(2x-1))(2x+7)-(2x+6)√(2x-1) = (2x+7)2 = (3x+2)(2x+7)-(2x+6)(2x-1) = 2×2+15x+20 (2x+7)2√(2x-1) (2x+7)2√(2x-1) 5.30. y’ = (3+1/(2√x))√(x2+2)-(3x+√x)x/√(x2+2) = x2+2 = (6√x+1)(x2+2)-2x√x(3x+√x) = 12√x+2-x2 2√x(x2+2)3/2 2√x(x2+2)3/2 5.31. y’ = (18×5+16×3-2x)√(1+x2)-x(3×6+4×4-x2-3)/√(1+x2) = 16×7+14×5+16×4+15×3 15+15×2 15(1+x2)3/2