search points attachment profile arrow left arrow right star heart verified symbols equation arrow-down question mark check menu accountancyadministrationagriculturalalgebraallarabicartart_musicbelarusbelarus_altbiologybusinesscatalachemistrychineseeconomicsegzamenglishentrepreneurshipenvironmentethicseuskarafirst_aidfrenchgalegogeographygeologygeometrygermangrammarhealthhistoryindia_langindonesian_langinformaticsitalianjapanesekazachkazach_altkoreanlanguagelatinlawlife_scienceliteraturelogicmathematicsmusicnigerian_langother_languagesotherspedagogicsphilosophyphysical_educationphysicspoliticspsychologyreligionrpa_langrussianrussian_altsciencesecurityskillssocial_sciencesociologyspanishstatisticstechnologytourismtrafficukrainianukrainian_altukrainian_literaturewos_civilisation accountancyadministrationagriculturalalgebraall_1arabicartart_music_2belarusbelarus_altbiologybusiness_2catalachemistry_1chineseeconomicsexam_3englishentrepreneurshipenvironment_2ethicseuskarasecurity_1frenchgalegogeography_4geology_4geometrygermangrammarhealthhistoryindia-langindonesian-langinformaticsitalianjapanesekazachAsset 230koreanlanguagelatinlawlife-scienceliteraturelogic_2mathematicsmusicnigerian-langotherlanguagesother_1pedagogicsphilosophyphysical_educationphysicspoliticspsychologyreligion_1rpa-langrussianrussian_altsciencesecurity_3_mskills_1allsocial_science_5_msociologyspanishstatisticstechnologytourismtrafficukrainianukrainian_altukrainian_literaturewos_civilisation
Автор - ViktorC

Найти производную данной функции и упростить ее:
y=tg(x/2) -ctg (x/2)

Ответ

Проверено экспертом

Автор - Artem112
y=tg frac{x}{2} -ctg frac{x}{2} <br />
\<br />
y'= frac{1}{2cos^2 frac{x}{2} } -(- frac{1}{2sin frac{x}{2} } )=<br />
 frac{1}{2cos^2 frac{x}{2} } +frac{1}{2sin ^2frac{x}{2} } =<br />
\<br />
= frac{sin ^2frac{x}{2}}{2sin ^2frac{x}{2}cos^2 frac{x}{2} } +frac{cos^2 frac{x}{2}}{2sin ^2frac{x}{2}cos^2 frac{x}{2} } =frac{sin^2frac{x}{2}+cos^2frac{x}{2}}{2sin ^2frac{x}{2}cos^2 frac{x}{2} } =frac{2}{4sin ^2frac{x}{2}cos^2 frac{x}{2} } =frac{2}{sin ^2x }

Ответы и объяснения

По всем вопросам пишите на - [email protected]
Сайт znanija.net не имеет отношения к другим сайтам и не является официальным сайтом компании.