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-x