tex] в точке x = 1 ...

June 2022 1 16 Report

Найдите угловой коэффициент касательной к графику функции y= In 8 [tex] log_{2} [/tex] x + [tex] x^{-1} [/tex] в точке x = 1

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Artem112

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$y= \ln 8 \log_{2} x + x^{-1} 
\\\
y`= \ln 8\cdot \frac{1}{x\ln2} - \frac{1}{x^2} = \frac{\ln 8}{x\ln2} - \frac{1}{x^2} 
\\\
y`(1)= \frac{\ln 8}{1\ln2} - \frac{1}{1^2} = \frac{3\ln 2}{\ln2} -1 = 3 -1 =2$

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ABC1234567 August 2022 | 0 Ответы

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