Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 53

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\[\textbf{а)}\ f(x) = e^{3x};\ \ x \in R\]

\[y = e^{u};\ \ u = 3x;\ \]

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

\[= e^{u} \cdot 3 = 3e^{3x}.\]

\[\textbf{б)}\ f(x) = e^{- 4x};\ \ x \in R\]

\[y = e^{u};\ \ u = - 4x;\ \]

\[f^{'}(x) = \left( e^{u} \right)_{u}^{'} \cdot ( - 4x)_{x}^{'} =\]

\[= e^{u} \cdot ( - 4) = - 4e^{- 4x}\text{.\ }\]

\[\textbf{в)}\ f(x) = e^{2x + 1};\ \ x \in R\]

\[y = e^{u};\ \ u = 2x + 1;\ \]

\[f^{'}(x) = \left( e^{u} \right)_{u}^{'} \cdot (2x + 1)_{x}^{'} =\]

\[= e^{u} \cdot 2 = 2e^{2x + 1}.\]

\[\textbf{г)}\ f(x) = e^{- 2x + 7};\ \ x \in R\]

\[y = e^{u};\ \ u = - 2x + 7;\ \]

\[f^{'}(x) = \left( e^{u} \right)_{u}^{'} \cdot ( - 2x + 7)_{x}^{'} =\]

\[= e^{u} \cdot ( - 2) = - 2e^{- 2x + 7}.\]

\[\textbf{д)}\ f(x) = 2^{5x};\ \ x \in R\]

\[y = 2^{u};\ \ u = 5x;\ \]

\[f^{'}(x) = \left( 2^{u} \right)_{u}^{'} \cdot (5x)_{x}^{'} =\]

\[= 2^{u} \cdot \ln 2 \cdot 5 = 5 \cdot 2^{5x}\ln 2.\]

\[\textbf{е)}\ f(x) = 6^{- 3x};\ \ x \in R\]

\[y = 6^{u};\ \ u = - 3x;\ \]

\[f^{'}(x) = \left( 6^{u} \right)_{u}^{'} \cdot ( - 3x)_{x}^{'} =\]

\[= 6^{u} \cdot \ln 6 \cdot ( - 3) = - 3 \cdot 6^{- 3x}\ln 6.\]

\[\textbf{ж)}\ f(x) = 4^{3x - 8};\ \ x \in R\]

\[y = 4^{u};\ \ u = 3x - 8;\ \]

\[f^{'}(x) = \left( 4^{u} \right)_{u}^{'} \cdot (3x - 8)_{x}^{'} =\]

\[= 4^{u} \cdot \ln 4 \cdot 3 = 3 \cdot 4^{3x - 8}\ln 4.\]

\[\textbf{з)}\ f(x) = 5^{- 4x + 1};\ \ x \in R\]

\[y = 5^{u};\ \ u = - 4x + 1;\ \]

\[f^{'}(x) = \left( 5^{u} \right)_{u}^{'} \cdot ( - 4x + 1)_{x}^{'} =\]

\[= 5^{u} \cdot \ln 5 \cdot ( - 4) =\]

\[= - 4 \cdot 5^{- 4x + 1}\ln 5.\]

\[\textbf{и)}\ f(x) = 2^{- 0,5x - 2};\ \ x \in R\]

\[y = 2^{u};\ \ u = - 0,5x - 2;\ \]

\[f^{'}(x) = \left( 2^{u} \right)_{u}^{'} \cdot ( - 0,5x - 2)_{x}^{'} =\]

\[= 2^{u} \cdot \ln 2 \cdot ( - 0,5) =\]

\[= - 0,5 \cdot 2^{- 0,5x - 2}\ln 2.\]