Question

Даю 35 баллов.
Помогите решить В2.

Answer 1

Не во всех примерах я доводила до конца преобразования

Answer 2

1.
$f'(x)=2x- \frac{1}{ x^{2} }= \frac{2x^3-1}{ x^{2} }$

2.
$f'(x)=-5x^4$

3.
$f'(x)=-6x^2+ \frac{6}{ \sqrt{x} }$

4.
$f(x)= \frac{7}{ \sqrt[4]{x} }- \frac{3}{x}=7*x^{- \frac{1}{4} }- \frac{3}{x} \\ \\ f'(x)=- \frac{7}{4x \sqrt{x} } + \frac{3}{x^2}= \frac{12-7 \sqrt{x} }{4x^2}$
5.
$f(x)=(x-6)x^3=x^4-6x^3 \\ \\ f'(x)=4x^3-18x^2$

6.
$f(x)= \sqrt{x} (x^2+ \frac{1}{ \sqrt{x} } )=x^{2.5}+1 \\ \\ f'(x)=2.5x^{1.5}=2.5x \sqrt{x}$

7.
$f'(x)= \frac{6}{2 \sqrt{6x+1} }*(x^4-5)+4x^3 \sqrt{6x+1}= \\ \\ = \frac{3(x^4-5)}{ \sqrt{6x+1} }+4x^3 \sqrt{6x+1}= \frac{3x^4-15+4x^3(6x+1)}{ \sqrt{6x+1} }= \\ \\ = \frac{3x^4-15+24x^4+4x^3}{ \sqrt{6x+1} }= \frac{27x^4+4x^3-15}{ \sqrt{6x+1} }$

8.
$f'(x)=( \frac{x}{3}+1 )^3+x*3( \frac{x}{3}+1 )^2* \frac{1}{3}= \\ \\ =( \frac{x}{3}+1 )^3+x( \frac{x}{3}+1 )^2=( \frac{x}{3}+1 )^2( \frac{x}{3}+1+x )= \\ \\ =( \frac{x}{3}+1 )^2( \frac{x+3+3x}{3} )= \frac{1}{3}(4x+3)( \frac{x}{3}+1 )^2$

9.
$f'(x)= \frac{2(3-2x)+2(2x+3)}{(3-2x)^2}= \frac{6-4x+4x+6}{(3-2x)^2}= \frac{12}{(3-2x)^2}$

10.
$f'(x)= \frac{3x^2(2x-3)-2x^3}{(2x-3)^2}= \frac{6x^3-9x^2-2x^3}{(2x-3)^2}= \frac{4x^3-9x^2}{(2x-3)^2}$

11.
$f'(x)= \frac{(4x^3+2x)(x+1)-(x^4+x^2+1)}{(x+1)^2}= \\ \\ = \frac{4x^4+2x^2+4x^3+2x-x^4-x^2-x}{(x+1)^2}= \\ \\ = \frac{3x^4+4x^3+x^2+2x-1}{(x+1)^2}$

12.
$f'(x)= \frac{2x^5(3x-2)-3( \frac{1}{3}x^6+2 )}{(3x-2)^2}= \frac{6x^6-4x^5-x^6-6}{(3x-2)^2}= \frac{5x^6-4x^5-6}{(3x-2)^2}$

13.
$f'(x)= \frac{15x^2(x-4)^2-5x^3*2(x-4)}{(x-4)^4}= \frac{15x^2(x^2-8x+16)-10x^4+40x^3}{(x-4)^4}= \\ \\ = \frac{15x^4-120x^3+240x^2-10x^4+40x^3}{(x-4)^4}= \frac{5x^4-80x^3+240x^2}{(x-4)^4}$

14.
$f'(x)= \frac{2x(x^3-x)+(x^2+1)(3x^2-1)}{(x^3-x)^2}= \frac{2x^4-2x^2+3x^4+3x^2-x^2-1}{(x^3-x)^2}= \\ \\ = \frac{5x^4-1}{(x^3-x)^2}$