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243_James Stewart Calculus 5 Edition Answers

# 243_James Stewart Calculus 5 Edition Answers - Stewart...

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f ( t )= 3 1+tan t =(1+tan t ) 1/3 ± f / ( t )= 1 3 (1+tan t ) ² 2/3 sec 2 t = sec 2 t 3 3 (1+tan t ) 2 y =cos ( a 3 + x 3 ) ± y / = ² sin ( a 3 + x 3 ) ³ 3 x 2 a 3 = ² 3 x 2 sin ( a 3 + x 3 ) y = a 3 +cos 3 x ± y / =3(cos x ) 2 ( ² sin x ) a 3 = ² 3sin x cos 2 x y = e ² mx ± y / = e ² mx d dx ( ² mx )= e ² mx ( ² m )= ² me ² mx y =4sec 5 x ± y / =4sec 5 x tan 5 x (5)=20sec 5 x tan 5 x g ( x )=(1+4 x ) 5 (3+ x ² x 2 ) 8 ± g / ( x ) =(1+4 x ) 5 ³ 8(3+ x ² x 2 ) 7 (1 ² 2 x )+(3+ x ² x 2 ) 8 ³ 5(1+4 x ) 4 ³ 4 =4(1+4 x ) 4 (3+ x ² x 2 ) 7 [2(1+4 x )(1 ² 2 x )+5(3+ x ² x 2 )] =4(1+4 x ) 4 (3+ x ² x 2 ) 7 [(2+4 x ² 16 x 2 )+(15+5 x ² 5 x 2 )] =4(1+4 x ) 4 (3+ x ² x 2 ) 7 (17+9 x ² 21 x 2 ) h ( t )=( t 4 ² 1) 3 ( t 3 +1) 4 ± h / ( t ) =( t 4 ² 1) 3 ³ 4( t 3 +1) 3 (3 t 2 )+( t 3 +1) 4 ³ 3( t 4 ² 1) 2 (4 t 3 ) =12 t 2 ( t 4 ² 1) 2 ( t 3 +1) 3 [( t 4 ² 1)+ t ( t 3 +1)]=12 t 2 ( t 4 ² 1) 2 ( t 3 +1) 3 (2 t 4 + t ² 1) y =(2 x ² 5) 4 (8 x 2 ² 5) ² 3 ± y / =4(2 x ² 5) 3 (2)(8 x 2 ² 5) ² 3 +(2 x ² 5) 4 ( ² 3)(8 x 2 ² 5) ² 4 (16 x ) =8(2 x ² 5) 3 (8 x 2 ² 5) ² 3 ² 48 x (2 x ² 5) 4 (8 x 2 ² 5) ² 4 y =( x 2 +1)( x 2 +2) 1/3 ± y / =2 x ( x 2 +2) 1/3 +( x 2 +1) 1 3 ( x 2 +2)
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