Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 832

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\[1)\ f(x) = e^{2x + 1} + 2x^{3}\]

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

\[= 2e^{2x + 1} + 2 \bullet 3x^{2} =\]

\[= 2e^{2x + 1} + 6x^{2}\]

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

\[f^{'}(x) = \left( e^{\frac{1}{2}x - 1} \right)^{'} - {(x - 1)^{\frac{1}{2}}\ }^{'} =\]

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

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

\[3)\ f(x) = e^{0,3x + 2} + \frac{1}{\sqrt{x}}\]

\[f^{'}(x) = \left( e^{0,3x + 2} \right)^{'} + \left( x^{- \frac{1}{2}} \right)^{'} =\]

\[= 0,3e^{0,3x + 2} - \frac{1}{2}x^{- \frac{3}{2}} =\]

\[= 0,3e^{0,3x + 2} - \frac{1}{2x\sqrt{x}}\]

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

\[f^{'}(x) = \left( e^{1 - x} \right)^{'} + \left( x^{- 3} \right)^{'} =\]

\[= - e^{1 - x} - 3x^{- 4}\]

\[5)\ f(x) = e^{x^{2}}\]

\[u = x^{2}\text{\ \ }f(u) = e^{u}:\]

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

\[= 2x \bullet e^{u} = 2x \bullet e^{x^{2}}\]

\[6)\ f(x) = e^{2x^{3}}\]

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

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

\[= 2 \bullet 3x^{2} \bullet e^{u} = 6x^{2} \bullet e^{2x^{3}}.\]