Pset4 - y = 1 & 2 x 2 : Its &rst...

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Ec115 Problem Set 4. The rules of di/erentiation. (i) y = 3 + 2 x ; y = c + 2 x; y = c + 2 x + bx 2 where b; c are any numbers. (ii) y = x (1 + x ); y = x (1 + 2 x 2 ); y = (1 + x + 2 x 2 )(1 3 x 3 ) : (iii) y = 1 x 1; y = x 1 (1 x ); y = 1 x x . (iv) y = x 1+ x ; y = x 2 1+ x : (v) y = (1 + 2 x ) 3 ; y = (1 + 2 x 2 ) 3 ; y = (1 + 2 x 3 ) 4 : (vi) y = x (1+2 x ) 2 ; y = x (1 x ) 2 If you need them, there are a lot more practice examples and questions (with answers) in Renshaw. 2. (a) Suppose output Q = 1 + 4 x + xy: Q with respect to x when ( i ) y = 0 ; ( ii ) y = 1; ( iii ) y = y 0 where y 0 is any number. On the same graph, draw each of these functions. How does a change in y shift this function? 3. Repeat question 2 but this time for production function Q = x 1 2 y 1 2 . 4. Consider the function
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Unformatted text preview: y = 1 & 2 x 2 : Its &rst derivative is dy dx = & 4 x: Finding where dy=dx = 0 requires solving dy dx = & 4 x = 0 : Doing this implies x = 0 : Now &nd where dy=dx = 0 for the following functions: (i) y = 1 & x 2 ; (ii) y = 1 & 2 x + x 2 ; y = x (1 & a & x ) & cx where a; c are any numbers. 5. By selling output q 1 ; &rm 1 makes pro&t & 1 = q 1 (1 & q 1 & q 2 ) & cq 1 where c; q 2 are any positive constants. Find q 1 where d& 1 =dq 1 = 0 : [If you have problems doing this, put c = 0 : 2 ; q 2 = 0 : 2 and answer the question ] : 1...
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