ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 199

\[1)\ y = 8\sqrt[4]{x} + 16e^{\frac{x}{2}};\]

\[y^{'}(x) = 8 \bullet \frac{1}{4}x^{- \frac{3}{4}} + 16 \bullet \frac{1}{2}e^{\frac{x}{2}};\]

\[y^{'}(x) = \frac{2}{\sqrt[4]{x^{3}}} + 8e^{\frac{x}{2}}.\]

\[2)\ y = \frac{9}{\sqrt[3]{x}} - \frac{1}{4}\sin{4x};\]

\[y^{'}(x) = 9 \bullet \left( - \frac{1}{3}x^{- \frac{4}{3}} \right) - \frac{1}{4} \bullet 4\cos{4x};\]

\[y^{'}(x) = - \frac{3}{x\sqrt[3]{x}} - \cos{4x}.\]

\[3)\ y = 3x\sqrt[3]{x} - 3\ln\frac{1}{x};\]

\[y^{'}(x) = 3 \bullet \frac{4}{3}x^{\frac{1}{3}} - 3 \bullet \left( - \frac{1}{x^{2}} \right) \bullet \left( 1\ :\frac{1}{x} \right);\]

\[y^{'}(x) = 4\sqrt[3]{x} + \frac{3}{x}.\]

\[4)\ y = \frac{1}{x\sqrt{x}} + 5\cos\frac{x}{5};\]

\[y^{'}(x) = - \frac{3}{2} \bullet x^{- \frac{5}{2}} + 5 \bullet \left( - \frac{1}{5}\sin\frac{x}{5} \right);\]

\[y^{'}(x) = - \frac{3}{2x^{2}\sqrt{x}} - \sin\frac{x}{5}.\]