Помогите решить срочно

Question 1

Question 2

1) (x^3-4)'=3x^2
2) (1/x + 2x)'= -1/x^2 + 2
3) (5x^5- $\sqrt{x}$ )' 25x^4 - 1/2* $\sqrt{x}$
4) (1/2 * x^2 + 4 $\sqrt{x}$ - 2/x)' = x + 2/ $\sqrt{x}$ + 2/x^2
5) ((3x+7)(7x^3+5x-4))'=3(7x^3+5x-4)+(21x^2+5)(3x+7)=21x^3+15x-12+63x^3+147x^2+15x+35= 84x^3+147x^2+30x+47
6) $( \frac{4x^2+8}{5-2x^3})' = 4* (\frac{x^2+2}{5-2x^3})'= 4* \frac{2x(5-2x^3)-(-6x^2)(x^2+2)}{(5-2x^3)^2}$ = $4* \frac{10x-4x^4+6x^4+12x^2}{(5-2x^3)^2} = 4* \frac{2x^2+12x^2+10x}{(5-2x^3)^2}$
1) (x^5+1)'=5x^4
2) (-1/x - 3x)'= 1/x^2-3
3) (4x^4 + $\sqrt{x}$ )= 16x^3 + 1/2* $\sqrt{x}$
4) (1/3 * x^3 - 2* $\sqrt{x}$ +5/x)'=x^2 - 1/ $\sqrt{x}$ - 5/x^2
5) ((5x-4)(2x^4-7x+1))'=5(2x^4-7x+1)+(8x^3-7)(5x-4)=10x^4-35x+5+40x^4-32x^3-35x+28=50x^4-32x^3-70x+33
6) $(\frac{x^3-7}{3-4x^4})'= \frac{3x^2(3-4x^4)-(-16x^3)(x^3-7)}{(3-4x^4)^2}= \frac{9x^2-12x^6+16x^6-112x^3}{(3-4x^4)^2}=[tex] \frac{4x^6-112x^3+9x^2}{(3-4x^4)^2}$

Abilay1411_zn · Answer 1 · 2018-05-27T02:47:18+0000

1) (x^3-4)'=3x^2
2) (1/x + 2x)'= -1/x^2 + 2
3) (5x^5- $\sqrt{x}$ )' 25x^4 - 1/2* $\sqrt{x}$
4) (1/2 * x^2 + 4 $\sqrt{x}$ - 2/x)' = x + 2/ $\sqrt{x}$ + 2/x^2
5) ((3x+7)(7x^3+5x-4))'=3(7x^3+5x-4)+(21x^2+5)(3x+7)=21x^3+15x-12+63x^3+147x^2+15x+35= 84x^3+147x^2+30x+47
6) $( \frac{4x^2+8}{5-2x^3})' = 4* (\frac{x^2+2}{5-2x^3})'= 4* \frac{2x(5-2x^3)-(-6x^2)(x^2+2)}{(5-2x^3)^2}$ = $4* \frac{10x-4x^4+6x^4+12x^2}{(5-2x^3)^2} = 4* \frac{2x^2+12x^2+10x}{(5-2x^3)^2}$
1) (x^5+1)'=5x^4
2) (-1/x - 3x)'= 1/x^2-3
3) (4x^4 + $\sqrt{x}$ )= 16x^3 + 1/2* $\sqrt{x}$
4) (1/3 * x^3 - 2* $\sqrt{x}$ +5/x)'=x^2 - 1/ $\sqrt{x}$ - 5/x^2
5) ((5x-4)(2x^4-7x+1))'=5(2x^4-7x+1)+(8x^3-7)(5x-4)=10x^4-35x+5+40x^4-32x^3-35x+28=50x^4-32x^3-70x+33
6) $(\frac{x^3-7}{3-4x^4})'= \frac{3x^2(3-4x^4)-(-16x^3)(x^3-7)}{(3-4x^4)^2}= \frac{9x^2-12x^6+16x^6-112x^3}{(3-4x^4)^2}=[tex] \frac{4x^6-112x^3+9x^2}{(3-4x^4)^2}$