Question

Нужно решение срочно даю 100 баллов,очень срочно!!!!!!!!!!!!!!!!!!!!!!!!!!​

Answer 1

Решение во вложении:

Answer 2

$1)\; \; (x^4+5)'=4x^3\\\\\\2)\; \; (7x-\frac{1}{x})'=7-(x^{-1})'=7-(-1\cdot x^{-2})=7+\frac{1}{x^2}\\\\\\3)\; \; (\sqrt{x}+3x^3)'=\frac{1}{2\sqrt{x}}+9x^2\\\\\\4)\; \; (\frac{1}{5}x^3+3\sqrt{x}+\frac{4}{x})'=\frac{1}{5}\cdot 3x^2+3\cdot \frac{1}{2\sqrt{x}}-\frac{4}{x^2}=\frac{3x^2}{5}+\frac{3}{2\sqrt{x}}-\frac{4}{x^2}$

$5)\; \; ((4x-6)(3x^5+2x-\sqrt5))'=\\\\=(4x-6)'(3x^5+2x-\sqrt5)+(4x-6)(3x^5+2x-\sqrt5)'=\\\\=4\, (3x^5+2x-\sqrt5)+(4x-6)(15x^4+2-0)=\\\\=12x^5+8x-4\sqrt5+60x^5+8x-90x^4-12=72x^5-90x^4+16x-12-4\sqrt5$

$6)\; \; \Big (\dfrac{x^3+5x}{-4x+x^2}\Big )'=\dfrac{(x^3+5x)'(-4x+x^2)-(x^3+5x)(-4x+x^2)'}{(-4x+x^2)^2}=\\\\=\dfrac{(3x^2+5)(-4x+x^2)-(x^3+5x)(-4+2x)}{(x^2-4x)^2}=\\\\=\dfrac{-12x^3+3x^4-20x+5x^2+4x^3-2x^4+20x-10x^2}{(x^2-4x)^2}=\\\\=\dfrac{x^4-8x^3-5x^2}{(x^2-4x)^2}=\dfrac{x^2(x^2-8x-5)}{x^2(x-4)^2}=\dfrac{x^2-8x-5}{(x-4)^2}$