Реферат: Решения к Сборнику заданий по высшей математике Кузнецова Л.А. - 2. Дифференцирование. Зад.5

Задача 5. Найти производную.

5.1.

(9x² +8x-1)(x+1)^1/2 – (3x³ +4x² -x-2)

y'=2/15* ___________________2(1+x)^1/2 =

1+x

= 2/15* (2x+2)(9x² +8x-1)-3x³ -4x² +x+2 =

2(x+1)^3/2

=2/15* 18x³ +16x² -2x+18x² +16x-2-3x³ -4x² +x+2 =

2(x+1)^3/2

= 2/15* 15x³ +30x² +15x =

2(x+1)^3/2

= x(x+1)² = x(x+1)^1/2

(x+1)^3/2

5.2.

3x³ *4x(x² +1)^1/2 +x(2x² -1) -9x² (2x² -1)(x² +1)^1/2

y'= (x² +1)^1/2 =

9x⁶

= 12x⁴ (x² +1)+3x⁴ (2x² -1)-9x² (2x² -1)(1+x² ) =

9x⁶ (x² +1)^1/2

= 12x⁴ +12x⁶ +6x⁶ -3x⁴ -18x⁴ -18x⁶ +9x² +9x⁴ =

9x⁶ (x² +1)^1/2

= 9x² = 1 .

9x⁶ (x² +1)^1/2 x⁴ (x² +1)^1/2

5.3.

y'= (4x³ -16x)(x² -4)-(x⁴ -8x² )2x = 4x⁵ -16x³ -16x³ +64x-2x⁵ +16x³ =

2(x² -4)² 2(x² -4)²

=2x⁵ -16x³ +64x =x(x² -4)² +16x = x+ 16x² .

2(x² -4)² (x² -4)² (x² -4)²

5.4.

(4x-1)√(2+4x) – 2(2x² -x-1)

y'= √(2+4x) = (4x-1)(2+4x)-4x² +x+1 =

3(2+4x) 3(2+4x)√(2+4x)

= 12x² +5x-1 .

3(2+4x)√(2+4x)

5. 5.

8x¹⁹ √(1+x⁸ )+ 4x¹⁹ (1+x⁸ ) – 12x¹¹ (1+x⁸ )^3/2

y'= √(1+x⁸ ) =

12x²⁴

= 12x¹⁹ (1+x⁸ )-12x¹¹ (1+x⁸ )² =

12x²⁴ √(1+x⁸ )

= x 11( x 16-2 x 8+1) = ( x 8-1)2 .

x²⁴ √(1+x⁸ ) x¹³ √(1+x⁸ )

5.6.

2x√(1-3x⁴ ) + 6 x ⁵ ­

y'= √(1-3 x ⁴ ) = 2 x (1-3 x ⁴ )+6 x ⁵ = x .

2(1-3x⁴ ) 2(1-3x⁴ )√(1-3x⁴ ) √(1-3x⁴ )³

5.7.

y= (2x(4+x² )√(4+x² )+3/2√(4+x² )*2x)x⁵ -(x² -6)(4+x² )√(4+x² )*5x⁴ =

120x¹⁰

= √(4+x² )(8x⁶ +2x⁸ +3x⁶ -20x⁶ -5x⁸ +30x⁶ +120x⁴ ) =

120x¹⁰

= √(4+x² )(7x² -x⁴ +40)

40x⁶

5.8.

y= 3/2√(x² -8)*2x⁴ -(x² -8)√(x² -8)*18x² =

6x⁶

√(x² -8)(x⁴ -6x⁴ +48x² ) = √(x² -8)(48-5x² )

3x⁶ 3x⁴

5.9.

9x³ (2+x³ )^2/3 -(4+3x³ )((2+x³ )^2/3 +2/3* 3x³ )

y'= (2+x³ )^1/3 =

x² (2+x³ )^4/3

= 9x³ (2+x³ )-(4+3x³ )(2+3x³ ) = 8 .

x² (2+x³ )^5/3 x² (2+x³ )^5/3

5.10.

y'= √(x)*(2(1+x^3/4 )*3/4x^5/4 -(1+x^3/4 )² *3/2*√(x)) =

3(1+x^3/4 )^2/3 *x^6/4

= √(x)(x^3/2 -1)

2x(1+x^3/2 )^2/3

5.11.

(6x⁵ +3x² )√(1-x³ ) + 3x² (x⁶ +x³ -2)

y' = 2√(1-x³ ) =

1-x³

=(2-2x³ )(6x⁵ +3x² )+3x⁸ +3x⁵ -6x² = (9x⁵ -9x⁸ ) = 9x⁵ .

2(1-x³ )^3/2 2(1-x³ )^3/2 2√(1-x³ )

5.12.

2x⁴ √(4+x² )+ x⁴ (x² -2) -3x² (x² -2)√(4+x² )

y'= √(4+x² ) =

24x⁶

= 2x⁴ (4+x² )+x⁴ (x² -2)-3x² (x² -2)(4+x² ) = 1

24x⁶ x⁴

5.13.

2x√(1+2x² )- 2x(1+x² )

y'= √(1+2x² ) = x(1+2x² )-x(1+x² ) = x^{3 .}

2(1+2x² ) (1+2x² )^3/2 (1+2x² )^3/2

5.14.

y'= ((3x+2)/(2√(x-1))+3√(x-1))x² -2x√(3x+2) =

4x⁴

= x² (3x+2)+6x² (x-1)-4x(x-1)(3x+2) = 9x³ -12x² +8x = 9x² -12x+8

4x² √(x-1) 4x² √(x-1) 4x√(x-1)

5.15.

y'= 3/2*√(1+x² )*2x⁴ -3x² (1+x² )^3/2 = √(1+x² )*(x⁴ -x² -x⁴ ) = -√(1+x² )

3x⁶ x⁶ x⁴

5.16.

(6x⁵ +24x² )√(8-x³ )+3x² (x⁶ +8x³ -128)

y'= 2√(8-x³ ) =

8-x³

= (16-2x³ )(6x⁵ +24x² )+3x² (x⁶ +8x³ -128) = 72x⁵ -9x⁸ = 9x⁵

2(8-x³ )^3/2 2(8-x³ )^3/2 2√(8-x³ )

5.17.

x² (x-2) +x² √(2x+3)-(2x² -4x)√(2x+3)

y'= √(2x+3) =

x⁴

= x² (x-2+2x+3)-(2x² -4x)(2x+3) = 3x² -x³ +12x = 3x-x² +12

x⁴ √(2x+3) x⁴ √(2x+3) x³ √(2x+3)

5.18.

y'=-2x⁵ √(x³ +1/x)+(1-x² )*1/5*(x³ +1/x)^4/5 *(3x² -1/x² )=1/5*(x³ +1/x)^4/5 (3x² -1/x² -3x⁴ +1)-2x(x³ +1/x)^1/5

5.19.

4x⁴ √(x² -3)+x⁴ (2x² +3) - 3x² (2x² +3)√(x² -3)

y' = √(x² -3) =

9x⁶

= 4x⁴ (x² -3)+x⁴ (2x² +3)-3x² (2x² +3)(x² -3) = 27x² = 3 .

9x⁶ √(x² -3) 9x⁶ √(x² -3) x⁴ √(x² -3)

5.20.

y'= (x² +5)^3/2 -3/2*(x-1)√(x² +5)*2x = √(x² +5)(5+3x-2x² )

(x² +5)³ (x² +5)³

5.21.

2x² √(x² -x)+(2x-1)(2x+1)x² -2x(2x+1)√(x² -x)

y'= √(x² -x) =

x⁴

= x² (2x² -2x+4x² -1)-(4x² +2x)(x² -x) = 2x² +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) = - 2x² -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 = -3x² -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)))-(2x² +x)√(x+1) =

(x² +x+1)²

= (3x+2)(x² +x+1)-(4x² +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

2-2x⁴ (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) = 2x² +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' = (18x⁵ +16x³ -2x)√(1+x² )-x(3x⁶ +4x⁴ -x² -3)/√(1+x² ) = 16x⁷ +14x⁵ +16x⁴ +15x³

15+15x² 15(1+x² )^3/2