ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 847

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\[1)\ f(x) = 2^{\cos x + 1}\]

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

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

\[f^{'}(x) = - \sin x \bullet 2^{u} \bullet \ln 2\]

\[f^{'}(x) = - 2^{\cos x + 1} \bullet \ln 2 \bullet \sin x.\]

\[2)\ f(x) = {0,5}^{1 + \sin x}\]

\[u = 1 + \sin x\ f(u) = {0,5}^{u}:\]

\[f^{'}(x) = \left( 1 + \sin x \right)^{'} \bullet \left( {0,5}^{u} \right)^{'}\]

\[f^{'}(x) = \cos x \bullet {0,5}^{u} \bullet \ln{0,5}\]

\[f^{'}(x) = - {0,5}^{1 + \sin x} \bullet \ln 2 \bullet \cos x.\]

\[3)\ f(x) = \cos\sqrt[3]{x + 2}\]

\[u = \sqrt[3]{x + 2}\ \ f(u) = \cos u:\]

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

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

\[f^{'}(x) = - \frac{\sin\sqrt[3]{x + 2}}{\sqrt[3]{(x + 2)^{2}}}.\]

\[4)\ f(x) = \sin\left( \ln x \right)\]

\[u = \ln x\text{\ f}(u) = \sin u:\]

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

\[f^{'}(x) = \frac{1}{x} \bullet \cos u\]

\[f^{'}(x) = \frac{\cos\left( \ln x \right)}{x}.\]