b The first derivative of 1 2 2 7 3 2 g x x x is 1 2 6 dg x x x dx Thus the

# B the first derivative of 1 2 2 7 3 2 g x x x is 1 2

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(b) The first derivative of 1 2 2 ( ) 7 3 2 g x x x = + is 1 2 ( ) 6 dg x x x dx = . Thus, the first derivative of ( ) g x when x = 4 would be 4 2 ( 7 ) 4 x dg x dx = = . 4. Product rule: ( ) ( ) ( ) g x h x k x = ( ) '( ) ( ) ( ) ( ) g x h x k x h x k x = + . for this question 3 ( ) ln( ) g x x x = ,where 3 ( ) h x x = and ( ) ln( ) k x x = thus, 2 3 2 2 ( ) ( ) ( ) ( ) ( ) 3 ln( ) (1/ ) 3 ln( ) g x h x k x h x k x x x x x x x x = + = + = + 5. Quotient rule: 2 ( ) ( ) ( ) '( ) '( ) ( ) ( ) : ( ) f y d g y f y g y f y g y dy g y = For this question: ( ) 3 2 '( ) 3 ( ) 5 7 '( ) 5 f y y f y g y y g y = + = = = Hence, 2 2 2 ( ) '( ) '( ) ( ) (5 7) 3 (3 2) 5 31 ( ) (5 7) 25 70 49 g y f y g y f y y y g y y y y × + × = = +
3 6. Chain rule: ( ) : df z df dy dz dy dz = × , For this question: 2 2 3 1 2 3 2 1 (2 12 ) 3 ( ) (2 12 ) (2 12 ) (2 24 ) df z z dy f z z z dy y z z z dz = + = + = + = + Hence: 2 2 2 3 2 3 ( ) 1 (2 24 )(2 12 ) : (2 12 ) (2 24 ) 3 3 df z df dy z z z z z z dz dy dz + + = × = + × + = Below is the graph of 2 10 y x + = ,we plot it as 1 5 2 y x = . The area under the line in the positive quadrant is the area of a right triangle with a base of 10 and a height of 5 10 5 / 2 25 × = Alternatively, we can use calculus, the area equals to the integration of y with

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