HW4solution-Part1

# HW4solution-Part1 - Math 16A Fall 2007 Professor Stankova...

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Math 16A, Fall 2007 Professor Stankova Félice Lê Homework 4 Solutions – Part 1 Section 1.6 1.6#4 3 3 1 y = x is equivalent to . Using the Power Rule, 1 3 ) 3 ( = x y 4 6 2 2 3 2 2 2 3 ) 9 ( ) x x = 3 2 3 1 9 9 ) 3 ( 9 3 ( 1 ) 3 ( ) 3 ( 1 x x x x x x dx d x dx dy = = = = . d Note that we have used both the Constant-Multiple Rule and the Power Rule in calculating ) 3 ( 3 x dx d . . 9 ) 3 ( 3 ) ( 3 ) 3 ( 2 2 3 3 x x x dx d x dx d = = = 1.6#8 Using the Sum Rule, we have . 1 ) 2 ( 4 2 3 dx d x dx d x dx d x dx d dx dy + + + = 2 2 3 3 12 ) 3 ( 4 4 4 x x x dx d x dx d = = = , by the Constant-Multiple and Power Rules. x x x dx d x dx d 4 ) 2 ( 2 2 ) 2 ( 2 2 = = = , by the Constant-Multiple and Power Rules. 1 = x dx d , by the Power Rule, and 0 1 = dx d , a special case of the Power Rule. Then . 1 4 12 2 + = x x dx dy 1.6#20 Again using the Sum Rule, . ) 1 ( 3 ) 1 ( 2 2 2 2 + + = x dx d x dx d dx dy x x x x x dx d x x dx d 4 4 ) 2 )( 1 ( 2 ) 1 ( ) 1 ( 2 ) 1 ( 3 2 2 2 2 2 + = + = + + = + , by the Power Rule. x x x x x dx d x x dx d x dx d 12 12 ) 2 )]( 1 ( 2 [ 3 ) 1 ( )] 1 ( 2 [ 3 ) 1 ( 3 ) 1 ( 3 3 2 2 2 2 2 2 2 = = = = , by the Constant-Multiple and Power Rules. Adding and combining like terms, we get . 8 16 12 12 4 4 3 3 3 x x x x x x dx dy = + + = 1.6#24 We can rewrite the function as . ) 1 ( 2 4 2 + = x y So using the Constant-Multiple Rule, 4 2 ) 1 ( 2 + = x dx d dx dy . 1

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Then using the Power Rule, 3 2 3 2 2 3 2 ) 1 ( 16 )] 2 ( ) 1 ( 4 [ 2 )] 1 ( ) 1 ( 4 [ 2 + = + = + + = x x x x x dx d x dx dy . 1.6#28 Rewrite the function as 2 1 2 ) 1 ( x y + = . Using the Power Rule, ). 2 ( ) 1 ( 2 1 ) 1 ( ) 1 ( 2 1 2 1 2 2 2 1 2 x x x dx d x dx dy + = + + = We can simplify this expression, giving 2 1 2 ) 1 ( + = x x dx dy or 2 1 x x dx dy + = or 2 1 2 ) 1 ( x x dx dy + = . 1.6#30 Rewrite the function as . 1 ) 5 2 ( + = x y Using the Power Rule, 2 2 2 ) 5 2 ( 2 ) 2 ( ) 5 2 ( 1 ) 5 2 ( ) 5 2 ( 1 + = + = + + = x x x dx d x dx dy or 2 ) 5 2 ( 2 + x . 1.6#36 is simply a constant. We know that if m is a constant, then 2 π m mx dx d = , so 2 2 ) ( dx d = = x dx dy . 1.6#38 Using the Power Rule, ) 1 ( ) 1 ( 1 2 x x dx d x x dx dy = . In order to find ) 1 ( x x dx d , we rewrite the expression as ) ( 1 x x dx d . Using the Sum Rule and then the Power Rule, we get 2 2 1 1 1 ) ) 1 ( ( 1 ) ( ) ( + = + = + = x x x dx d x dx d x x dx d , or 2 1 1 x + . Putting this expression into dx dy and simplifying, we now have 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ) 1 ( 1 ) 1 ( 1 1 1 1 1 1 ) 1 1 ( ) 1 ( 1 + = + = + = + = + = x x x x x x x x x x x x x x x x dx dy . 1.6#44 The tangent line will have an equation of the form b mx y + = , where dx dy m = .
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HW4solution-Part1 - Math 16A Fall 2007 Professor Stankova...

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