ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 848

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

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

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

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

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

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

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

\[= \cos x \bullet \frac{1}{3} \bullet u^{- \frac{2}{3}} = \frac{\cos x}{3\sqrt[3]{u^{2}}} =\]

\[= \frac{\cos x}{3\sqrt[3]{\sin^{2}x}}.\]

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

\[u = \cos x\ \text{\ f}(u) = \sqrt[4]{u}:\]

\[f^{'}(x) = \left( \cos x \right)^{'} \bullet \left( u^{\frac{1}{4}} \right)^{'} =\]

\[= - \sin x \bullet \frac{1}{4} \bullet u^{- \frac{3}{4}} = - \frac{\sin x}{4\sqrt[4]{u^{3}}} =\]

\[= - \frac{\sin x}{4\sqrt[4]{\cos^{3}x}}.\]

\[4)\ f(x) = \sqrt{\log_{2}x}\]

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

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

\[= \frac{1}{x \bullet \ln 2} \bullet \frac{1}{2\sqrt{u}} =\]

\[= \frac{1}{2x \bullet \ln 2 \bullet \sqrt{\log_{2}x}}.\]