ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 235

\[1)\ f(x) = e^{\sin^{2}x + \sin x};\text{\ \ \ }x_{0} = \pi:\]

\[f(\pi) = e^{\sin^{2}\pi + \sin\pi} = e^{0^{2} + 0} =\]

\[= e^{0} = 1;\]

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

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

\[f^{'}(\pi) = (2 \bullet 0 \bullet 1 - 1) \bullet e^{0^{2} + 0} =\]

\[= - e^{0} = - 1;\]

\[y = 1 - 1 \bullet (x - \pi) = \pi + 1 - x.\]

\[Ответ:\ \ y = \pi + 1 - x.\]

\[2)\ f(x) = \frac{1}{x^{2}\ }\sin\frac{\pi x^{2}}{2};\ x_{0} = 1:\]

\[f^{'}(1) = \frac{1}{1} \bullet \sin\frac{\pi}{2} = 1 \bullet 1 = 1;\]

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

\[= - \frac{2}{x^{3}} \bullet \sin\frac{\pi x^{2}}{2} + \frac{1}{x^{2}} \bullet \frac{\pi}{2} \bullet 2x \bullet \cos\frac{\pi x^{2}}{2};\]

\[f^{'}(1) =\]

\[= - \frac{2}{1} \bullet \sin\frac{\pi}{2} + \frac{1}{1} \bullet \pi \bullet \cos\frac{\pi}{2} = - 2;\]

\[y = 1 - 2(x - 1) = 3 - 2x.\]

\[Ответ:\ \ y = 3 - 2x.\]