Задача Кузнецов Дифференцирование 10-24

Условие задачи

Найти производную.

$y={\frac {8}{3}}\cdot \operatorname {cth} {2x}-{\frac {1}{3\operatorname {ch} {x}\cdot \operatorname {sh} ^{3}{x}}}$

Решение

$y'=\left({\frac {8}{3}}\cdot \operatorname {cth} {2x}-{\frac {1}{3\operatorname {ch} {x}\cdot \operatorname {sh} ^{3}{x}}}\right)'=\left({\frac {8}{3}}\cdot {\frac {1}{\operatorname {th} {2x}}}-{\frac {1}{3\operatorname {ch} {x}\cdot \operatorname {sh} ^{3}{x}}}\right)'=$

$=-{\frac {8}{3}}\cdot {\frac {1}{\operatorname {th} ^{2}{2x}}}\cdot {\frac {1}{\operatorname {ch} ^{2}{2x}}}\cdot 2+{\frac {1}{3\operatorname {ch} ^{2}{x}\cdot \operatorname {sh} ^{6}{x}}}\cdot \left(\operatorname {ch} {x}\cdot \operatorname {sh} ^{3}{x}\right)'=$

$=-{\frac {16}{3\operatorname {sh} ^{2}{2x}}}+{\frac {1}{3\operatorname {ch} ^{2}{x}\cdot \operatorname {sh} ^{6}{x}}}\cdot \left(\operatorname {sh} {x}\cdot \operatorname {sh} ^{3}{x}+\operatorname {ch} {x}\cdot 3\operatorname {sh} ^{2}{x}\cdot \operatorname {ch} {x}\right)=$

$=-{\frac {16}{3\operatorname {ch} ^{2}{x}\cdot \operatorname {sh} ^{2}{x}}}+{\frac {\operatorname {sh} ^{2}{x}+3\operatorname {ch} ^{2}{x}}{3\operatorname {ch} ^{2}{x}\cdot \operatorname {sh} ^{4}{x}}}={\frac {-16\operatorname {sh} ^{2}{x}+\operatorname {sh} ^{2}{x}+3\operatorname {ch} ^{2}{x}}{3\operatorname {ch} ^{2}{x}\cdot \operatorname {sh} ^{4}{x}}}=$

$={\frac {-5\operatorname {sh} ^{2}{x}+\operatorname {ch} ^{2}{x}}{\operatorname {ch} ^{2}{x}\cdot \operatorname {sh} ^{4}{x}}}={\frac {-5\operatorname {sh} ^{2}{x}+1+\operatorname {sh} ^{2}{x}}{\operatorname {ch} ^{2}{x}\cdot \operatorname {sh} ^{4}{x}}}={\frac {1-4\operatorname {sh} ^{2}{x}}{\operatorname {ch} ^{2}{x}\cdot \operatorname {sh} ^{4}{x}}}$