Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 73

\[\boxed{\mathbf{73}\mathbf{.}}\]

\[\textbf{а)}\ y = \arccos\text{πx};\ \ x \in \left( - \frac{1}{\pi};\frac{1}{\pi} \right)\]

\[g^{'}(x) = \left( \arccos x \right)^{'} = - \frac{1}{\sqrt{1 - x^{2}}};\]

\[f^{'}(x) = - \frac{1}{\sqrt{1 - \left( \text{πx} \right)^{2}}} \cdot \pi =\]

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

\[\textbf{б)}\ y = arctg\ x^{3};\ \ x \in R\]

\[g^{'}(x) = \left( \text{arctg\ x} \right)^{'} = \frac{1}{1 + x^{2}};\]

\[f^{'}(x) = \frac{1}{1 + \left( x^{3} \right)^{2}} \cdot 3x^{2} = \frac{3x^{2}}{1 + x^{6}}.\]

\[\textbf{в)}\ y = \arcsin{5x};\ \ x \in \left( - \frac{1}{5};\frac{1}{5} \right)\]

\[g^{'}(x) = \left( \arcsin x \right)^{'} = \frac{1}{\sqrt{1 - x^{2}}};\]

\[f^{'}(x) = \frac{1}{\sqrt{1 - (5x)^{2}}} \cdot 5 =\]

\[= \frac{5}{\sqrt{1 - 25x^{2}}}.\]

\[\textbf{г)}\ y = (arcctg\ 3x)^{5};\ \ \ x \in R\]

\[g^{'}(x) = \left( \text{arcctg\ x} \right)^{'} = - \frac{1}{1 + x^{2}};\]

\[f^{'}(x) = 5(arcctg\ 3x)^{4} \cdot\]

\[\cdot \left( - \frac{1}{1 + (3x)^{2}} \right) \cdot 3 =\]

\[= - \frac{15(arcctg\ 3x)^{4}}{1 + 9x^{2}}.\]

\[\textbf{д)}\ y = \arccos( - 2x);\ \ \]

\[x \in \left( - \frac{1}{2};\frac{1}{2} \right)\]

\[g^{'}(x) = \left( \arccos x \right)^{'} =\]

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

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

\[= \frac{2}{\sqrt{1 - 4x^{2}}}.\]

\[\textbf{е)}\ y = \left( \arcsin{4x} \right)^{5};\ \ x \in \left( - \frac{1}{4};\frac{1}{4} \right)\]

\[g^{'}(x) = \left( \arcsin x \right)^{'} = \frac{1}{\sqrt{1 - x^{2}}};\]

\[f^{'}(x) = 5\left( \arcsin{4x} \right)^{4} \cdot\]

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

\[= \frac{20\left( \arcsin{4x} \right)^{4}}{\sqrt{1 - 16x^{2}}}.\]

## Параграф 5. Применение производной