# ep2b - 92.131(t 2 1 8(t 4 1 7 Calculus1 Chain Rule II 1 s(t...

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92.131 Calculus1 Chain Rule II 1) 7 4 8 2 ) 1 ( ) 1 ( ) ( + + = t t t s 2) 10 6 2 ] 3 ) 3 [( ) ( + + = t t t s 3) ) (sec tan ) ( 4 t t f = 4) 8 3 2 ] 1 cos [ ) ( + + = x x x f 5) ) sin( 2 2 x e y x = 6) 5 4 3 ] 1 ) 2 [( ) ( + = t t t s 7) ) tan(csc ) ( x x f = 8) 1 sin ) ( 3 + = x x f 9) ) 1 cos( ) ( 4 + = x x f 10) 5 3 4 ] 3 ) 3 [( ) ( + + = t t t s 11) 4 ) ( x e x f = 12) x e x f tan ) ( = 13) ) sec( 9 x e y = 14) x e y ln = 15) 5 ) ( x e x f = 16) ) ln(tan x y = 17) 5 4 ) 1 ( ) ( + = x e x f 18) ) tan( 2 2 ) 1 sin( x e x y + = 19) ) ln( ) ( 3 x x x f + = 20) ) ln( ) ( cos sin x x e e x f = 21) x x e y ln = 22) ) sin( 2 x e y = 23) Find the second derivative of ] 4 cos 4 [sin 4 x x e y x + = 24) )] ln[sin( 3 3 x e y x = 25) [ ] 100 2 4 2 ) 1 ( 1 ln ) ( + + + = x x x x g

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92.131 Calculus1 Chain Rule II Solutions 1) [ ] 8 4 3 5 7 2 ) 1 ( 4 7 3 ) 1 ( 4 ) ( + + + = t t t t t t s 2) 9 6 2 5 2 5 2 9 6 2 ] 3 ) 3 [( ) 3 )( 180 120 ( )] 3 2 ( ) 3 ( 6 [ ] 3 ) 3 [( 10 ) ( + + + + = + + + + = t
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ep2b - 92.131(t 2 1 8(t 4 1 7 Calculus1 Chain Rule II 1 s(t...

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