Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 846

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\[1)\ f(x) = \ln\sqrt{x - 1}\]

\[u = \sqrt{x - 1}\ f(u) = \ln u:\]

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

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

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

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

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

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

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

\[f^{'}(x) = \frac{1}{2} \bullet (3 + x)^{- \frac{1}{2}} \bullet e^{u}\]

\[f^{'}(x) = \frac{e^{\sqrt{3 + x}}}{2 \bullet \sqrt{3 + x}}.\]

\[3)\ f(x) = \ln\left( \cos x \right)\]

\[u = \cos x\ f(u) = \ln u:\]

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

\[f^{'}(x) = - \sin x \bullet \frac{1}{u}\]

\[f^{'}(x) = - \frac{\sin x}{\cos x}\]

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

\[4)\ f(x) = \ln\left( \sin x \right)\text{\ \ }\]

\[u = \sin x\ f(u) = \ln u:\]

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

\[f^{'}(x) = \cos x \bullet \frac{1}{u}\]

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

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