Hàm số y = tanx có đạo hàm cấp 2 bằng :
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Lời giải:
Báo saita có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaacaaIXaaabaGaae4yaiaab+gacaqGZbWaaWba % aSqabeaacaaIYaaaaOGaamiEaaaaaaa!3D8C! y' = \frac{1}{{{\rm{co}}{{\rm{s}}^2}x}}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaaga % Gaeyypa0JaeyOeI0YaaSaaaeaacaaIYaGaae4yaiaab+gacaqGZbGa % amiEamaabmaabaGaeyOeI0Iaae4CaiaabMgacaqGUbGaamiEaaGaay % jkaiaawMcaaaqaaiaabogacaqGVbGaae4CamaaCaaaleqabaGaaGin % aaaakiaadIhaaaGaeyypa0ZaaSaaaeaacaaIYaGaae4CaiaabMgaca % qGUbGaamiEaaqaaiaabogacaqGVbGaae4CamaaCaaaleqabaGaaG4m % aaaakiaadIhaaaaaaa!52EF! y'' = - \frac{{2{\rm{cos}}x\left( { - {\rm{sin}}x} \right)}}{{{\rm{co}}{{\rm{s}}^4}x}} = \frac{{2{\rm{sin}}x}}{{{\rm{co}}{{\rm{s}}^3}x}}\)
