ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 879

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\[1)\ y = \cos^{2}{3x}\]

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

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

\[= - 3\sin{3x} \bullet 2u =\]

\[= - 6\sin{3x} \bullet \cos{3x} = - 3\sin{6x}.\]

\[2)\ y = \sin x \bullet \cos x + x\]

\[y^{'}(x) =\]

\[= \left( \sin x \right)^{'} \bullet \cos x + \sin x \bullet \left( \cos x \right)^{'} + (x)^{'}\]

\[y^{'} = \cos^{2}x - \sin^{2}x + 1 =\]

\[= \cos^{2}x - \sin^{2}x + \cos^{2}x + \sin^{2}x =\]

\[= 2\cos^{2}x.\]

\[3)\ y = \left( x^{3} + 1 \right) \bullet \cos{2x}\]

\[y^{'}(x) =\]

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

\[y^{'} =\]

\[= 3x^{2} \bullet \cos{2x} + \left( x^{3} + 1 \right) \bullet \left( - 2\sin{2x} \right) =\]

\[= 3x^{2} \bullet \cos{2x} - 2\left( x^{3} + 1 \right) \bullet \sin{2x}.\]

\[4)\ y = \sin^{2}\frac{x}{2} = \frac{1 - \cos x}{2}\]

\[y^{'}(x) = \left( \frac{1}{2} \right)^{'} - \frac{1}{2} \bullet \left( \cos x \right)^{'} =\]

\[= 0 - \frac{1}{2} \bullet \left( - \sin x \right) = \frac{1}{2} \bullet \sin x.\]

\[5)\ y = (x + 1) \bullet \sqrt[3]{x^{2}}\]

\[y^{'}(x) =\]

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

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

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

\[= \frac{3x + 2x + 2}{3\sqrt[3]{x}} = \frac{5x + 2}{3 \bullet \sqrt[3]{x}}.\]

\[6)\ y = \sqrt[3]{x - 1} \bullet \left( x^{4} - 1 \right)\]

\[y^{'}(x) =\]

\[= {(x - 1)^{\frac{1}{3}}}^{'} \bullet \left( x^{4} - 1 \right) + \sqrt[3]{x - 1} \bullet \left( x^{4} - 1 \right)^{'} =\]

\[= \frac{1}{3} \bullet (x - 1)^{- \frac{2}{3}} \bullet \left( x^{4} - 1 \right) + \sqrt[3]{x - 1} \bullet 4x^{3} =\]

\[= \frac{x^{4} - 1}{3 \bullet \sqrt[3]{(x - 1)^{2}}} + 4x^{3} \bullet \sqrt[3]{x - 1} =\]

\[= \frac{x^{4} - 1 + 12x^{3}(x - 1)}{3 \bullet \sqrt[3]{(x - 1)^{2}}} =\]

\[= \frac{x^{4} - 1 + 12x^{4} - 12x^{3}}{3 \bullet \sqrt[3]{(x - 1)^{2}}} =\]

\[= \frac{13x^{4} - 12x^{3} - 1}{3 \bullet \sqrt[3]{(x - 1)^{2}}}.\]