# ans_ex_2 - A M A 2 1 1 A n s w e r s t o E x e r c i s e 2...

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Unformatted text preview: A M A 2 1 1 A n s w e r s t o E x e r c i s e 2 ( 1 ) F o r a n y ∆ x n = , t h e c o m p o u n d a n g l e f o r m u l a i m p l i e s c o s ( x + ∆ x ) − c o s x ∆ x = c o s x b c o s ∆ x − 1 ∆ x B − s i n x b s i n ∆ x ∆ x B = − 2 c o s x s i n ∆ x 2 ¡ s i n ∆ x 2 ∆ x 2 ¢ − s i n x b s i n ∆ x ∆ x B . Y o u m a y t h e n a l l o w ∆ x → . ( i ) d d x t a n x = s e c 2 x , ( i i ) d d x c o t x = 1 t a n 2 x × d t a n x d x = 1 s i n 2 x , ( i i i ) d d x s e c x = t a n x s e c x , ( i v ) d d x c s c x = − c o t x c s c x , ( v ) d d x p x 2 c o s x x 3 + x + 1 P = ( x 3 + x + 1 ) ( 2 x c o s x − x 2 s i n x ) − x 2 c o s x ( 3 x 2 + 1 ) ( x 3 + x + 1 ) 2 . ( 2 ) W e h a v e l i m ∆ x → + f ( 1 + ∆ x ) − f ( 1 ) ∆ x = l i m ∆ x → + 2 ( 1 + ∆ x ) − 3 − ( − 1 ) ∆ x = 2 a n d l i m ∆ x → − f ( 1 + ∆ x ) − f ( 1 ) ∆ x = l i m ∆ x → − ( 1 + ∆ x ) − 2 − ( − 1 ) ∆ x = 1 . H e n c e f ′ ( 1 ) d o e s n o t e x i s t . ( 3 ) I f ∆ x > , t h e n f ( + ∆ x ) − f ( ) ∆ x = ∆ x 2 3 − ∆ x = 1 ∆ x 1 3 . T h u s l i m ∆ x → + f ( + ∆ x ) − f ( ) ∆ x = + ∞ a n d l i m ∆ x → − f ( + ∆ x ) − f ( ) ∆ x = − ∞ . T h e r e f o r e , f ′ ( ) d o e s n o t e x i s t . ( 4 ) W e u s e t h e c h a i n r u l e t o o b t a i n f ′ ( x ) = − 1 σ 3 √ 2 π ( x − μ ) e − 1 2 σ 2 ( x − μ ) 2 . T h u s f ′ ( x ) = i f a n d o n l y i f x − μ = . 1 ( 5 ) I f f ′ ( ) e x i s t s , t h e n a = 2 a n d b = 5 . ( 6 ) W e h a v e ( a ) 2 f ( a ) f ′ ( a ) ; ( b ) f ′ ( a ) ; ( c ) 2 f ′ ( a ) ; ( d ) 1 2 f ′ ( a ) . ( 7 ) W e h a v e ( a ) − e − x s i n ( x l n x ) + e − x [ 1 + l n x ] c o s ( x l n x ) ; ( b ) 4 e 2 x ( e 2 x + 1 ) 2 ; ( c ) 4 x x 4 − 1 ; ( d ) 2 x + 3 ( x 2 + 3 x − 1 ) l n ( x 2 + 3 x − 1 ) . ( 8 ) ∆ V = 1 5 π ( r + ∆ r ) 2 − 1 5 π r 2 = 1 5 π £ 2 r ∆ r + ( ∆ r ) 2 ¤ a n d d V = 3 π r ∆ r . T h e r e f o r e , ∆ V − d V = 1 5 π ( ∆ r ) 2 . ( 9 ) W e h a v e h = f ( a ) = 1 2 a + 6 . W h e n a = 1 5 a n d ∆ a = . 1 , w e h a v e h = 1 4 a n d h − h = ∆ h = f ( a + ∆ a ) − f ( a ) = 1 2 a + ∆ a − 1 2 a = 1 2 1 5 + . 1 − 1 2 1 5 ≃ − . 5 3 2 9 8 ....
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## This note was uploaded on 04/08/2012 for the course AMA 211 taught by Professor Chanchunwah during the Spring '12 term at Hong Kong Polytechnic University.

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ans_ex_2 - A M A 2 1 1 A n s w e r s t o E x e r c i s e 2...

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