M2_Bk1_Sol_Ch05_E.pdf

# M2_Bk1_Sol_Ch05_E.pdf - 5 New Progress in Senior...

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© Hong Kong Educational Publishing Co. 134 New Progress in Senior Mathematics Module 2 Book 1 (Extended Part) Solution Guide 5 5 pp.178 – 182 p.178 Let , q p n = where p and q are integers and . 0 q Hence p q p n x x x ) ( 4 1 = = Now let , 4 1 x u = we have . p u y = ) 2 ....... ( .......... ) 1 ....... ( .......... 1 1 = = q p qu du dx pu du dy 1 1 1 1 ) ( ) ( ) ( = = = = = n q p q q p n q n p nx x q p u q p x dx d qu x dx d pu du dx dx dy du dy p.182 (b) dx dy y 2 3 (c) 2 2 y dx dy xy + (d) xy dx dy x 2 2 + (e) 3 2 3 y dx dy xy + (f) y x dx dy x 2 3 3 + pp.169 – 184 5.1 (a) ) ( 4 1 4 6 6 x dx d x dx d dx dy = = p.169 2 3 6 4 1 5 5 x x = = (b) ) ( 4 ) 4 ( 6 6 = = x dx d x dx d dx dy 7 7 24 ) 6 ( 4 = = x x 5.2 ) 4 4 6 ( 3 4 + = x x dx d dx dy p.170 4 3 3 4 3 4 12 16 ) ( 4 ) ( 4 0 ) 4 ( ) 4 ( ) 6 ( = + = + = x x x dx d x dx d x dx d x dx d dx d 5.3 5 5 4 5 7 7 8 x x x x x y + = p.171 5 1 2 7 8 + = x x x 6 2 6 2 5 1 5 1 2 35 8 2 ) 5 ( 7 ) ( 8 2 ) ( 7 ) ( 8 2 ) 7 ( ) 8 ( ) ( + = + = + = + = x x x x x x x dx d x dx d x x dx d x dx d x dx d dx dy 5.4 dx dy p.172 5 2 6 6 2 2 1 4 ) 2 )( 3 ( ) 1 2 )( 1 2 ( ) 1 2 ( ) 3 ( ) 3 ( ) 1 2 ( 2 2 2 2 2 2 + + = + + + = + + + + = + + + + + = x x x x x x x x x x dx d x x x x dx d x 5.5 ) ( x f p.172 ) 1 3 5 )( 3 10 ( ) 10 3 )( 1 ( ) 1 ( ) 3 10 ( ) 3 10 ( ) 1 ( 2 4 3 2 3 5 3 5 3 3 3 5 + + + + + = + + + + + = x x x x x x x x x x x dx d x x x x dx d x x x 24 ) 1 3 5 )( 3 10 1 ( ) 10 3 )( 1 1 1 1 ( ) 1 ( = + + + + + = f 5.6 2 2 2 2 ) 3 ( ) 3 ( 10 ) 10 ( ) 3 ( + + + = x x dx d x x dx d x dx dy p.173 2 2 2 2 2 2 ) 3 ( ) 3 ( 10 ) 3 ( ) 2 ( 10 ) 10 )( 3 ( + = + + = x x x x x x Differentiation (1)

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135 © Hong Kong Educational Publishing Co. Differentiation (1) 5.7 4 4 4 2 2 + + + = x x x y p.174 4 4 1 2 + + = x x 2 2 2 2 ) 4 ( ) 4 ( ) 4 ( ) 4 ( ) 4 ( 0 + + + + = x x dx d x x dx d x dx dy 2 2 2 2 2 2 ) 4 ( 8 4 ) 4 ( ) 2 )( 4 ( ) 1 )( 4 ( + + = + + = x x x x x x x 5.8 Let , 4 u y = where . 5 5 2 + = x x u p.177 3 2 3 ) 5 5 )( 5 2 ( 4 ) 5 2 )( 4 ( + = = = x x x x u dx du du dy dx dy 5.9 ) 11 3 ( ] ) 11 3 ( 6 [ 4 7 = x dx d x dx dy p.178 7 7 ) 11 3 ( 72 ) 3 ( ) 11 3 ( 24 = = x x 5.10 (a) 2 ) 1 ( ) 1 ( ) ( 1 + + + = x x dx d x x dx d x dx dy p.179 2 3 2 3 2 1 ) 1 ( 2 2 ) 1 ( 2 ) 1 ( 2 1 ) 1 ( 1 2 1 1 ) 1 ( ) 1 ( 2 1 ) 1 ( 1 + + = + + = + + + = + + + + = x x x x x x x x x x x dx d x x x (d) ] ) 5 ( [ ] ) 5 ( [ 2 1 6 2 1 6 + + + + = x x dx d x x dx dy 2 1 6 5 5 2 1 6 5 2 1 6 ] ) 5 ( [ 2 ) 5 ( 6 1 ] ) 5 ( 6 1 [ ] ) 5 ( [ 2 1 ) 5 ( ) 5 ( 6 1 ] ) 5 ( [ 2 1 + + + + = + + + + = + + + + + = x x x x x x x dx d x x x 5.11 Differentiate both sides of the equation with p.182 respect to x , we have 2 3 2 3 2 3 2 4 8 2 2 ) 8 ( 0 0 2 8 ) 0 ( ) 1 2 ( x y xy dx dy xy dx dy x y xy dx dy x dx dy y dx d y x y dx d = = = + = 5.12 x x y y 2 3 2 2 = + p.183 xy y x dx dy y x dx dy xy x dx dy y dx dy y x x y xy x y x y 4 3 ) 3 ( 2 2 6 ) 4 3 ( 0 6 3 2 2 2 0 2 3 2 ) ( 2 3 2 2 2 2 2 2 3 2 2 2 + = + = = + + = + + = 5.13 Differentiate both sides of the equation with p.184 respect to x , we have ) ...( 2 2 3 2 3 ) 2 ( 0 3 2 2 2 2 2 2 2 2 2 2 2 + = = + = + + + xy x y xy x dx dy y xy x dx dy xy x x y dx dy xy xy dx dy x Substituting x = 2, 5 = y into ), ( we have 16 7 ) 5 )( 2 ( 2 ) 2 ( ) 5 ( ) 5 )( 2 ( 2 ) 2 ( 3 2 2 2 ) 5 , 2 ( = + = dx dy
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