# ep2a - 5 sin 5 3 cos 3 − = ′ 16 x x y tan sec 2 = ′...

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

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92.131 Calculus 1 Chain Rule I Solutions 1) ) cos( 2 ) ( 2 x x x f = 2) 9 2 ) 3 ( 20 ) ( + = t t t s 3) ) ( cos sin 4 ) ( 3 t t t f = 4) ) ( sec 3 ) ( 3 2 2 x x x f = 5) 2 2 x xe y = 6) 4 4 3 ) 1 ( 20 ) ( + = t t t s 7) x x x x f sec 2 tan sec ) ( = 8) x x x f 2 cos ) ( = 9) x e x y sin cos = 10) ) ( csc ) ( 2 x x e e x f = 11) 1 2 ) ( 2 + = x x x f 12) x e x x x f csc cot csc ) ( = 13) x e x x y sec tan sec = 14) x e y 4 4 = 15) x x x f
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Unformatted text preview: 5 sin 5 3 cos 3 ) ( − = ′ 16) x x y tan sec 2 = ′ 17) 1 3 4 4 ) ( + = ′ x e x x f 18) 4 4 ) 1 sin( 4 ) 1 cos( 2 2 3 2 x x e x x e x x y + + + = ′ 19) x x f cot ) ( = ′ 20) x x f cos ) ( = ′ 21) x x e x y ln ) ln 1 ( + = 22) x x y 2 sec ) cos(tan = 23) ] 3 cos 3 [sin 18 3 x x e y x + = ′ ′ 24) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ′ x x e y x 1 ] ln[ 3 3 3...
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ep2a - 5 sin 5 3 cos 3 − = ′ 16 x x y tan sec 2 = ′...

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