Помогите пожалуйста! С 1 заданием (2,5,6,7,8) Буду благодарна!. Created by syutkinamari. matematika-ru

Помогите пожалуйста! С 1 заданием (2,5,6,7,8) Буду благодарна!...

syutkinamari @syutkinamari

July 2022 1 7 Report

Помогите пожалуйста!
С 1 заданием (2,5,6,7,8)
Буду благодарна!

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NNNLLL54

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$2)\; \; y=\sqrt[3]{x}\cdot arctgx\; \; ,\; \; \; (u\cdot v)'=u'v+uv'\\\\y'=\frac{1}{3}\cdot x^{- \frac{2}{3}}\cdot arctgx+\sqrt[3]{x}\cdot \frac{1}{1+x^2}\\\\5)\; \; y=ln\frac{x^2}{1-x^2}\; \; ,\; \; \; (lnu)'=\frac{1}{u}\cdot u'\\\\y'=\frac{1-x^2}{x^2}\cdot \frac{2x\cdot (1-x^2)-x^2\cdot (-2x)}{(1-x^2)^2}=\frac{1-x^2}{x^2}\cdot \frac{2x-2x^3+2x^3}{(1-x^2)^2}=\frac{1}{x^2}\cdot \frac{2x}{1-x^2}=\frac{2}{x(1-x^2)}$

$6)\; \; y=10^{3-lg2x}=10^3\cdot 10^{-lg2x}=10^3\cdot 10^{lg(2x)^{-1}}=10^3\cdot (2x)^{-1}\\\\y=10^3\cdot \frac{1}{2x}=500\cdot \frac{1}{x}\\\\y'=500\cdot (\frac{1}{x})'=500\cdot \frac{-1}{x^2}=-\frac{500}{x^2}\\\\7)\; \; y=arccos\, e^{\frac{x^2}{2}}\; \; ,\; \; (arccos\, u)'=-\frac{1}{\sqrt{1-u^2}}\cdot u'\\\\y'=-\frac{1}{\sqrt{1-e^{x^2}}}\cdot e^{\frac{x^2}{2}}\cdot \frac{1}{2}\cdot 2x=-\frac{x\cdot e^{\frac{x^2}{2}}}{\sqrt{1-e^{x^2}}}$

$8)\; \; y=log_7(cos\sqrt{1+x}\, )\; \; ,\; \; (log_7\,u)'=\frac{1}{u\, ln7} \cdot u'\\\\y'=\frac{1}{cos\sqrt{1+x}\; \cdot ln7}\cdot (cos\sqrt{1+x})'=\Big [\; (cosu)'=-sinu\cdot u'\; \Big ]=\\\\=\frac{1}{ln7\cdot cos\sqrt{1+x}}\cdot (-sin\sqrt{1+x}\, )\cdot (\sqrt{1+x})'=\Big [\; (\sqrt{u})'=\frac{1}{2\sqrt{u}}\cdot u'\; \Big ]=\\\\=-\frac{sin\sqrt{1+x}}{ln7\cdot cos\sqrt{1+x}}\cdot \frac{1}{2\sqrt{1+x}}\cdot 1=-\frac{tg\sqrt{1+x}}{2\, \cdot \, ln7\cdot \sqrt{1+x}}$

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syutkinamari August 2022 | 0 Ответы

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syutkinamari August 2022 | 0 Ответы

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