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Question 1

Question 2

Решите задачу:

$1. \displaystyle (x^4+5)'=(x^4)'+5'=\bf {4x^3}$
$2. \displaystyle (7x- \frac{1}{x} )'=(7x)'-( \frac{1}{x} )'=7-(- \frac{1}{x^2})=\bf{7+ \frac{1}{x^2} }$
$3. \displaystyle (\sqrt x+3x^3)'=(\sqrt x)'+(3x^3)'=\bf{ \frac{1}{2 \sqrt x} +9x^2}$
$4. \displaystyle ( \frac{x^3}{5} +3 \sqrt x + \frac{4}{x})'= ( \frac{x^3}{5})'+(3 \sqrt x )'+( \frac{4}{x})' \boxed{=}$
$\displaystyle ( \frac{x^3}{5} )'= \frac{(x^3)' \cdot 5- x^3 \cdot 5'}{5^2} = \frac{3x^2 \cdot 5-x^3 \cdot 0}{25}= \frac{15x^2}{25}= \frac{3x^2}{5}$
$\displaystyle (3 \sqrt x)'=3' \cdot \sqrt x + 3 \cdot (\sqrt x)'=0+ 3 \cdot \frac{1}{2 \sqrt x}= \frac{3}{2 \sqrt x}$
$\displaystyle ( \frac{4}{x})'= \frac{4'x-4x' }{x^2}= -\frac{4}{x^2}$
$\displaystyle \boxed{=}\bf{ \frac{3x^2}{5}+ \frac{3}{2 \sqrt x}- \frac{4}{x^2}}$
$5. \displaystyle ((4x-6)(3x^5+2x- \sqrt 5))'=(4x-6)'(3x^5+2x- \sqrt 5)+$
$+(4x-6)(3x^5+2x- \sqrt 5)'=4(3x^5+2x- \sqrt 5)+(4x-6)(15x^4+2)=$
$=12x^5+8x-4 \sqrt 5 +60x^5+8x-90x^4-12=\bf{72x^5-90x^4+16x-}$
$\bf{-12-4 \sqrt 5}=2(36x^5-45x^4+8x-6-2 \sqrt 5)$
$\displaystyle 6. ( \frac{x^3+5x}{-4x+x^2} )' = \frac{(x^3+5x)'(-4x+x^2)-(x^3+5x)(-4x+x^2)'}{(-4x+x^2)^2 }=$
$\displaystyle = \frac{(3x^2+5)(-4x+x^2)-(x^3+5x)(-4+2x)}{(-4x+x^2)^2} =$
$\displaystyle = \frac{-12x^3+3x^4-20x+5x^2-(-4x^3+2x^4-20x+10x^2)}{(-4x+x^2)^2}=$
$\displaystyle = \frac{-12x^3+3x^4-20x+5x^2+4x^3-2x^4+20x-10x^2}{(-4x+x^2)^2}=$
$\displaystyle = \frac{x^4-8x^3-5x^2}{x^4-8x^3+16x^2}= \frac{x^2(x^2-8x-5)}{x^2(x^2-8x-16)}=\bf{ \frac{x^2-8x-5}{(x-4)^2} }$