Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 853

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\[1)\ f(x) = e^{2x} \bullet \ln(2x - 1)\]

\[= 2e^{2x} \bullet \ln(2x - 1) + e^{2x} \bullet \frac{2}{2x - 1} =\]

\[= 2e^{2x} \bullet \left( \ln(2x - 1) + \frac{1}{2x - 1} \right).\]

\[e^{2x} \bullet \ln(2x - 1) = 0\]

\[\ln{(2x - 1}) = 0\]

\[\ln(2x - 1) = \ln 1\]

\[2x - 1 = 1\]

\[2x = 2\]

\[x = 1.\]

\[f^{'}(1) = 2e^{2} \bullet \left( \ln(2 - 1) + \frac{1}{2 - 1} \right) =\]

\[= 2e^{2} \bullet \left( \ln 1 + 1 \right) = 2e^{2}.\]

\[Ответ:\ \ 2e^{2}.\]

\[2)\ f(x) = \frac{\sin x - \cos x}{\sin x}\]

\[f^{'}(x) = \frac{1}{\sin^{2}x}\]

\[\frac{\sin x - \cos x}{\sin x} = 0\]

\[1 - ctg\ x = 0\]

\[\text{ctg\ x} = 1\]

\[x = arcctg\ 1 + \pi n = \frac{\pi}{4} + \pi n.\]

\[f^{'}\left( \frac{\pi}{4} + \pi n \right) = \frac{1}{\sin^{2}\left( \frac{\pi}{4} + \pi n \right)} =\]

\[= \frac{1}{\left( \sin\frac{\pi}{4} \right)^{2}} = 1\ :\left( \frac{1}{\sqrt{2}} \right)^{2} =\]

\[= 1\ :\frac{1}{2} = 2.\]

\[Ответ:\ \ 2.\]