ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 881

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\[1)\ f(x) = \log_{2}\left( x^{3} - x^{2} + 1 \right)\]

\[u = x^{3} - x^{2} + 1\ f(u) = \log_{2}u:\]

\[f^{'}(x) = \left( x^{3} - x^{2} + 1 \right)^{'} \bullet \left( \log_{2}u \right)^{'} =\]

\[= \left( 3x^{2} - 2x \right) \bullet \frac{1}{u \bullet \ln 2} =\]

\[= \frac{3x^{2} - 2x}{(x^{3} - x^{2} + 1) \bullet \ln 2}.\]

\[2)\ f(x) = \left( \log_{2}x \right)^{3}\]

\[u = \log_{2}x\ f(u) = u^{3}:\]

\[f^{'}(x) = \left( \log_{2}x \right)^{'} \bullet \left( u^{3} \right)^{'} =\]

\[= \frac{1}{x \bullet \ln 2} \bullet 3u^{2} = \frac{3 \bullet \left( \log_{2}x \right)^{2}}{x \bullet \ln 2} =\]

\[= \frac{3 \bullet \ln^{2}x}{x \bullet \ln^{3}2}.\]

\[3)\ f(x) = \sin\left( \log_{3}x \right)\]

\[u = \log_{3}x\ f(u) = \sin u:\]

\[f^{'}(x) = \left( \log_{3}x \right)^{'} \bullet \left( \sin u \right)^{'} =\]

\[= \frac{1}{x \bullet \ln 3} \bullet \cos u = \frac{\cos\left( \log_{3}x \right)}{x \bullet \ln 3}.\]

\[4)\ f(x) = \cos 3^{x}\]

\[u = 3^{x}\ f(u) = \cos u:\]

\[f^{'}(x) = \left( 3^{x} \right)^{'} \bullet \left( \cos u \right)^{'} =\]

\[= 3^{x} \bullet \ln 3 \bullet \left( - \sin u \right) =\]

\[= - 3^{x} \bullet \ln 3 \bullet \sin 3^{x}.\]