Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 303

\[1)\ f(x) = \sin^{2}x;\]

\[f^{'}(x) = 2\sin x \bullet \cos x = \sin{2x};\]

\[f^{''}(x) = 2\cos{2x}.\]

\[2)\ f(x) = x^{3} \bullet \sin x;\]

\[f^{'}(x) = 3x^{2} \bullet \sin x + x^{3} \bullet \cos x;\]

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

\[= x\left( 6 - x^{2} \right) \bullet \sin x + 6x^{2} \bullet \cos x.\]

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

\[f^{'}(x) = 4x^{3} + 3 \bullet 2x - 1 + 0 =\]

\[= 4x^{3} + 6x - 1;\]

\[f^{''}(x) = 4 \bullet 3x^{2} + 6 - 0 = 12x^{2} + 6.\]

\[4)\ f(x) = x^{4} - 3x^{3} + 5x + 6;\]

\[f^{'}(x) = 4x^{3} - 3 \bullet 3x^{2} + 5 + 0 =\]

\[= 4x^{3} - 9x^{2} + 4;\]

\[f^{''}(x) = 4 \bullet 3x^{2} - 9 \bullet 2x + 0 =\]

\[= 12x^{2} - 18x.\]

\[5)\ f(x) = e^{\sin x};\]

\[f^{'}(x) = e^{\sin x} \bullet \cos x\]

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

\[= e^{\sin x} \bullet \cos x \bullet \cos x - e^{\sin x} \bullet \sin x =\]

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

\[6)\ f(x) = \ln\left( x^{2} + 1 \right);\]

\[f^{'}(x) = \frac{1}{x^{2} + 1} \bullet 2x =\]

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

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

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

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