Решебник по алгебре 11 класс Никольский Параграф 4. Производная Задание 58

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

\[f^{'}(x) = \cos x^{2} \cdot 2x = 2x\cos x^{2}.\]

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

\[f^{'}(x) = - \sin x^{4} \cdot 4x^{3} =\]

\[= - 4x^{3}\sin x^{4}.\]

\[\textbf{в)}\ f(x) = tg\ x^{3};\ \ \]

\[x^{3} \neq \frac{\pi}{2} + \pi k;\ \ x \neq \sqrt[3]{\frac{\pi}{2} + \pi k}\]

\[f^{'}(x) = \frac{1}{\cos^{2}x^{3}} \cdot 3x^{2} = \frac{3x^{2}}{\cos^{2}x^{3}}.\]

\[\textbf{г)}\ f(x) = ctg\ \left( x^{5} \right);\ \ \]

\[x^{5} \neq \pi k;\ \ x \neq \sqrt[5]{\text{πk}}\]

\[f^{'}(x) = - \frac{1}{\sin^{2}x^{5}} \cdot 5x^{4} =\]

\[= - \frac{5x^{4}}{\sin^{2}x^{5}}\text{.\ }\]

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

\[f^{'}(x) = 2\sin x \cdot \cos x = \sin{2x}.\]

\[\textbf{е)}\ f(x) = \left( \cos x \right)^{4};\ \ x \in R\]

\[f^{'}(x) = - \sin x \cdot 4\cos^{3}x =\]

\[= - 4\cos^{3}x \cdot \sin x.\]

\[\textbf{ж)}\ f(x) = \left( \text{tgx} \right)^{3};\ \ x \neq \frac{\pi}{2} + \pi k\]

\[f^{'}(x) = \frac{1}{\cos^{2}x} \cdot 3tg^{2}x = \frac{3tg^{2}x}{\cos^{2}x}\text{.\ }\]

\[\textbf{з)}\ f(x) = \left( \text{ctg\ x} \right)^{5};\ \ x \neq \pi k\]

\[f^{'}(x) = - \frac{1}{\sin^{2}x} \cdot 5ctg^{4}x =\]

\[= - \frac{5ctg^{4}x}{\sin^{2}x}.\]