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Varvara2001 @Varvara2001

July 2022 1 12 Report

Вычислите производную функций:
а) y = [tex]-2x^{-2} +1[/tex]
б) y = [tex]\frac{1}{2} x^{-2}[/tex]
в) y = - [tex]\frac{2}{x^3} [/tex]
г) y = [tex]3x^\frac{4}{3} [/tex]
д) y =[tex] \frac{2}{3} x^\frac{1}{3}[/tex]
е) y = [tex] \frac{1}{2\sqrt{x}} + x[/tex]
ж) y = [tex]\frac{3}{\sqrt[3]{x}} + \sqrt{x}[/tex]
з) y = [tex]\frac{1}{\sqrt[x]{x}} - 4[/tex]

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NNNLLL54

Verified answer

$1)\; \; y=-2x^{-2}+1\; ,\; \; \; y'=-2(-2)x^{-3}=\frac{4}{x^3}\\\\2)\; \; y=\frac{1}{2}x^{-2}\; ,\; \; y'= \frac{1}{2}\cdot (-2)\cdot x^{-3}=-x^{-3}\\\\3)\; \; y=-\frac{2}{x^3}\; ,\; \; y'=-\frac{-2\cdot 3x^2}{x^6}=\frac{6}{x^4}\; \; \; \; [\, (\frac{k}{u})'=\frac{-k\cdot u'}{u^2},\; k=const\, ]\\\\4)\; \; y=3x^{4/3}\; ,\; \; \; y'=3\cdot \frac{4}{3}\cdot x^{\frac{1}{3}}=4\sqrt[3]{x}\\\\5)\; \; y=\frac{2}{3}\cdot x^{1/3}\; ,\; \; y'=\frac{2}{3}\cdot \frac{1}{3}\cdot x^{-\frac{2}{3}}=\frac{2}{9\sqrt[3]{x^2}}$

$6)\; \; y=\frac{1}{2\sqrt{x}}+x\; ,\; \; y'=\frac{1}{2}\cdot (-\frac{1}{2})\cdot x^{-\frac{3}{2}}+1=- \frac{1}{4\sqrt{x^3}}+1\\\\7)\; \; y=\frac{3}{\sqrt[3]{x}}+\sqrt{x}\; ,\\\\y'=3\cdot (-\frac{1}{3})\cdot x^{-\frac{4}{3}}+\frac{1}{2}\cdot x^{-\frac{1}{2}}=-\frac{1}{\sqrt[3]{x^4}}+\frac{1}{2\sqrt{x}}\\\\8)\; \; y=\frac{1}{\sqrt[x]{x}}-4=x^{1/x}-4\\\\\\y=x^{1/x}\; \; \to \; \; lny=ln(x^{1/x})\; ,\; \; \Big (lny\Big )'=\Big (ln(x^{1/x})\Big )'\\\\\frac{y'}{y}=(\frac{1}{x}\cdot lnx)'\; \; \to \; \; y'=y\cdot (-\frac{1}{x^2}\cdot lnx+\frac{1}{x^2})=x^{1/x}\cdot \frac{1}{x^2}(1-lnx)$

$\Big (\frac{1}{\sqrt[x]{x}}-4\Big )'=x^{\frac{1}{x}}\cdot \frac{1}{x^2}\cdot (1-lnx)-0=\frac{\sqrt[x]{x}}{x^2}\cdot lnx$

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