O gabarito serve para você confirmar se acertou o processo de derivação. Caso sua resposta seja diferente do gabarito, examine sua solução de forma a detectar erros sutis que levaram a essa diferença.
Aqui estão as respostas para as questões:
� ′ ( � ) = 6 � f ′ ( x ) = 6 x � ′ ( � ) = 1 2 � + 6 � 2 g ′ ( x ) = 2 x 1 + 6 x 2 ℎ ′ ( � ) = − 1 � 2 + � � h ′ ( x ) = − x 2 1 + e x � ′ ( � ) = 2 � 3 + 3 � 2 − 2 � 2 f ′ ( x ) = x 2 2 x 3 + 3 x 2 − 2 � ′ ( � ) = 1 � + cos ( � ) g ′ ( x ) = x 1 + cos ( x ) ℎ ′ ( � ) = � � ( 2 − � 2 ) � 4 h ′ ( x ) = x 4 e x ( 2 − x 2 ) � ′ ( � ) = 16 ( 2 � 3 + 4 � − 1 ) 3 ( 6 � 2 + 4 ) f ′ ( x ) = 16 ( 2 x 3 + 4 x − 1 ) 3 ( 6 x 2 + 4 ) � ′ ( � ) = 3 � + 1 3 � 2 + 2 � + 1 g ′ ( x ) = 3 x 2 + 2 x + 1 3 x + 1 ℎ ′ ( � ) = − sin ( � ) ( � 2 + 1 ) − 2 � cos ( � ) ( � 2 + 1 ) 2 h ′ ( x ) = ( x 2 + 1 ) 2 − s i n ( x ) ( x 2 + 1 ) − 2 x c o s ( x ) � ′ ( � ) = 6 � + 5 3 � 2 + 5 � + 2 f ′ ( x ) = 3 x 2 + 5 x + 2 6 x + 5 � ′ ( � ) = 2 � 2 � cos ( � ) − � 2 � sin ( � ) g ′ ( x ) = 2 e 2 x cos ( x ) − e 2 x sin ( x ) ℎ ′ ( � ) = 3 sec 2 ( � ) − 2 tan ( � ) � 4 h ′ ( x ) = x 4 3 s e c 2 ( x ) − 2 t a n ( x ) � ′ ( � ) = − 1 2 � � − � − � f ′ ( x ) = − 2 x x 1 − e − x � ′ ( � ) = 0 g ′ ( x ) = 0 ℎ ′ ( � ) = 3 � 2 − 2 � 3 − 2 � + 1 h ′ ( x ) = x 3 − 2 x + 1 3 x 2 − 2 � ′ ( � ) = � � ( 1 − � ) 2 � 3 / 2 f ′ ( x ) = 2 x 3/2 e x ( 1 − x ) � ′ ( � ) = 3 sec 2 ( 3 � ) g ′ ( x ) = 3 sec 2 ( 3 x ) ℎ ′ ( � ) = 2 � 4 + 2 � − 2 ( � 3 + 2 � ) 2 h ′ ( x ) = ( x 3 + 2 x ) 2 2 x 4 + 2 x − 2 � ′ ( � ) = � � − 1 2 � − 1 / 2 � � 2 � � � f ′ ( x ) = 2 x e x e x − 2 1 x − 1/2 e x � ′ ( � ) = 4 � − 3 2 � 2 − 3 � + 1 g ′ ( x ) = 2 x 2 − 3 x + 1 4 x − 3 ℎ ′ ( � ) = 2 � � � 2 cos ( � ) − � � 2 sin ( � ) h ′ ( x ) = 2 x e x 2 cos ( x ) − e x 2 sin ( x ) � ′ ( � ) = 2 tan ( � ) sec 2 ( � ) f ′ ( x ) = 2 tan ( x ) sec 2 ( x ) � ′ ( � ) = � cos ( � ) − sin ( � ) cos 2 ( � ) g ′ ( x ) = c o s 2 ( x ) x c o s ( x ) − s i n ( x ) ℎ ′ ( � ) = − 2 � ( � 2 − 1 ) 2 h ′ ( x ) = ( x 2 − 1 ) 2 − 2 x � ′ ( � ) = � � + � � � 2 � f ′ ( x ) = 2 x e x + x e x � ′ ( � ) = cos ( � ) sin ( � ) g ′ ( x ) = s i n ( x ) c o s ( x ) ℎ ′ ( � ) = � � ( cos ( � ) − 2 sin ( � ) ) cos 4 ( � ) h ′ ( x ) = c o s 4 ( x ) e x ( c o s ( x ) − 2 s i n ( x )) � ′ ( � ) = sec 2 ( � ) � + tan ( � ) 2 � f ′ ( x ) = x s e c 2 ( x ) + 2 x t a n ( x ) � ′ ( � ) = � cos ( � ) − sin ( � ) ( � 2 + 1 ) ( � 2 + 1 ) 2 g ′ ( x ) = ( x 2 + 1 ) 2 x c o s ( x ) − s i n ( x ) ( x 2 + 1 ) ℎ ′ ( � ) = � � 2 � h ′ ( x ) = 2 x e x � ′ ( � ) = 6 � 2 − 8 � 2 � 3 − 4 � 2 + 1 f ′ ( x ) = 2 x 3 − 4 x 2 + 1 6 x 2 − 8 x � ′ ( � ) = − 2 sin ( 2 � ) g ′ ( x ) = − 2 sin ( 2 x ) ℎ ′ ( � ) = − 2 � cos ( � ) − sin ( � ) ( � 2 − 1 ) ( � 2 − 1 ) 2 h ′ ( x ) = ( x 2 − 1 ) 2 − 2 x c o s ( x ) − s i n ( x ) ( x 2 − 1 ) � ′ ( � ) = 1 + � � 2 � � f ′ ( x ) = x 2 1 + x e x � ′ ( � ) = � ( 2 sec 2 ( 2 � ) + tan ( 2 � ) ) g ′ ( x ) = x ( 2 sec 2 ( 2 x ) + tan ( 2 x )) ℎ ′ ( � ) = 2 � 2 � − 4 � � 2 � � 3 h ′ ( x ) = x 3 2 e 2 x − 4 x e 2 x � ′ ( � ) = � � 2 � � f ′ ( x ) = 2 e x e x � ′ ( � ) = − sin ( � ) cos ( � ) g ′ ( x ) = c o s ( x ) − s i n ( x ) ℎ ′ ( � ) = − 4 � cos ( � ) − sin ( � ) ( � 2 + 4 ) ( � 2 − 4 ) 2 h ′ ( x ) = ( x 2 − 4 ) 2 − 4 x c o s ( x ) − s i n ( x ) ( x 2 + 4 ) � ′ ( � ) = tan ( � ) + � sec 2 ( � ) � � f ′ ( x ) = e x t a n ( x ) + x s e c 2 ( x ) � ′ ( � ) = − sin ( � ) � cos ( � ) g ′ ( x ) = − sin ( x ) e c o s ( x ) ℎ ′ ( � ) = � 4 − 4 � 3 − 3 � 2 − 1 ( � 4 − 1 ) 2 h ′ ( x ) = ( x 4 − 1 ) 2 x 4 − 4 x 3 − 3 x 2 − 1 � ′ ( � ) = cos ( � ) 2 sin ( � ) f ′ ( x ) = 2 s i n ( x ) c o s ( x ) � ′ ( � ) = cos ( � ) sin ( � ) g ′ ( x ) = s i n ( x ) c o s ( x ) ℎ ′ ( � ) = − 2 � cos ( � ) − sin ( � ) ( � 2 + 4 ) ( � 2 + 4 ) 2 h ′ ( x ) = ( x 2 + 4 ) 2 − 2 x c o s ( x ) − s i n ( x ) ( x 2 + 4 ) � ′ ( � ) = ln ( � ) + 1 � cos 2 ( � ) + tan ( � ) ln ( � ) f ′ ( x ) = x c o s 2 ( x ) l n ( x ) + 1 + tan ( x ) ln ( x ) � ′ ( � ) = 9 � 2 − 6 � + 4 2 2 � 3 − 3 � 2 + 4 � − 1 g ′ ( x ) = 2 2 x 3 − 3 x 2 + 4 x − 1 9 x 2 − 6 x + 4 ℎ ′ ( � ) = sec 2 ( � ) � tan ( � ) h ′ ( x ) = sec 2 ( x ) e t a n ( x ) � ′ ( � ) = � � + ln ( � ) � � 2 � � f ′ ( x ) = 2 e x e x + l n ( x ) e x � ′ ( � ) = cos ( � ) − sin ( � ) sin ( � ) + cos ( � ) g ′ ( x ) = s i n ( x ) + c o s ( x ) c o s ( x ) − s i n ( x )