Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 248

\[1)\ y = \frac{1}{5}x^{2} + 2\ln x - \cos x;\]

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

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

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

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

\[= \frac{1}{2}x - e^{x} + 2\cos x.\]

\[3)\ y = 15\sqrt[5]{x} + e^{x} - 6\ tg\ x;\]

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

\[= 15 \bullet \frac{1}{5}x^{- \frac{4}{5}} + e^{x} - 6 \bullet \frac{1}{\cos^{2}x} =\]

\[= \frac{3}{\sqrt[5]{x^{4}}} + e^{x} - \frac{6}{\cos^{2}x}.\]

\[4)\ y = 6\sqrt[6]{x} - \ln x + \frac{1}{3}\cos x;\]

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

\[= 6 \bullet \frac{1}{6}x^{- \frac{5}{6}} - \frac{1}{x} + \frac{1}{3} \bullet \left( - \sin x \right) =\]

\[= \frac{1}{\sqrt[6]{x^{5}}} - \frac{1}{x} - \frac{1}{3}\sin x.\]

\[5)\ y =\]

\[= x^{2}(x - 1) + 3\sin x + 4\ ctg\ x;\]

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

\[= 3x^{2} - 2x + 3\cos x - \frac{4}{\sin^{2}x}.\]

\[6)\ y =\]

\[= x(x + 2)^{2} + 2\ln x - 3\cos x;\]

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

\[= 3x^{2} + 8x + 4 + \frac{2}{x} + 3\sin x.\]

\[7)\ y = (x - 1)(x + 2) + e^{x} - \ln x;\]

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

\[= (x + 2) + (x - 1) + e^{x} - \frac{1}{x} =\]

\[= 2x + 1 + e^{x} - \frac{1}{x}.\]

\[8)\ y =\]

\[= (x + 3)(2x - 1) + e^{x} - \sin x;\]

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

\[= (2x - 1) + 2(x + 3) + e^{x} - \cos x;\]

\[y^{'}(x) = 4x + 5 + e^{x} - \cos x.\]