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

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+x3/4
)*3/4x5/4
-(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.

(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
)
= x3 .

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)
=

= 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)
=

= 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

Обсуждение:

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

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