Question

решить Вычисление производных

Answer 1

$1)\ \ \ (x^2+x)'=7x^6+1\\\\(x^{-2}+\sqrt{x})'=(x^{-2}+x^{\frac{1}{2}})'=-2x^{-3}+\dfrac{1}{2}\, x^{-\frac{1}{2}}=-\dfrac{2}{x^3}+\dfrac{1}{2\sqrt{x}}\\\\(-4x+x^4)'=-4+4x^3\\\\(x^9+\dfrac{1}{x})'=(x^9+x^{-1})'=9x^8-x^{-2}=9x^8-\dfrac{1}{x^2}\\\\(x^{-1}-x^2+1)'=-x^{-2}-2x+0\\\\(6x-3+x^5)'=6+5x^4\\\\(4+\pi +\sqrt7)'=4'+\pi '+(\sqrt7)'=0+0+0=0\\\\(x^{10}+x-3)'=10x^9+1\\\\(x^7-3x+2)'=7x^6-3\\\\(x^{-5}+\sqrt{x}-\pi )'=-5x^{-6}+\dfrac{1}{2\sqrt{x}}-0$

$2)\ \ (3x^4)'=3\cdot 4x^3=12x^3\\\\(-2x^7)'=-2\cdot 7x^6=-14x^6\\\\(-7x^{-3})'=-7\cdot (-3)x^{-4}=21x^{-4}\\\\(1,5x^4)'=1,5\cdot 4x^3=6x^3\\\\(6\sqrt{x})'=6\cdot \dfrac{1}{2\sqrt{x}}\\\\\Big(\dfrac{3}{x}\Big)'=(3\cdot x^{-1})'=3\cdot (-1)\cdot x^{-2}=-\dfrac{3}{x^2}\\\\(-2\sqrt{x})'=-2\cdot \dfrac{1}{2\sqrt{x}}=-\dfrac{1}{\sqrt{x}}\\\\(3x^{-5})'=-15x^{-6}\\\\\Big(-\dfrac{1}{2x}\Big)'=\Big(-\dfrac{1}{2}\cdot x^{-1}\Big)'=-\dfrac{1}{2}\cdot x^{-2}=-\dfrac{1}{2x^2}$

$(\sqrt{2x})'=\sqrt2\cdot (\sqrt{x})'=\sqrt{2}\cdot \dfrac{1}{2\sqrt{x}}=\dfrac{1}{\sqrt2\cdot \sqrt{x}}=\dfrac{1}{\sqrt{2x}}$

Answer 2

$5.\\1)x^6+1\\2)-2x^{-3}+\frac{1}{2\sqrt{x}}\\3)-4+4x^3\\4)9x^8-\frac{1}{x^2} \\5)-\frac{1}{x^2}-2x\\6)6+5x^4\\7)0\\8)10x^9+1\\9)7x^6-3\\10)-5x^{-6}+\frac{1}{2\sqrt{x}}$

$6.\\1)12x^3\\2)-14x^6\\3)21x^{-4}\\4)6x^3\\5)\frac{3}{\sqrt{x}}\\6)-\frac{3}{x^2}\\7)-\frac{1}{\sqrt{x}}\\8)-15x^{-6}\\9)\frac{1}{2x^2}\\10)\frac{1}{\sqrt{2x}}$