Unformatted text preview: 4q )/2  [1/(2p)]/2 U = 4/2 +(7z 4q )/2  (1/4)(1/p) âˆ‚ U/ âˆ‚ p = (1/4) Â· âˆ‚ âˆ‚ p 1 p (note: in this partial derivative, treat "z" and "q" as constants) âˆ‚ U/ âˆ‚ p = (1/4) Â· ( 29 1 p pâˆ‚ âˆ‚ âˆ‚ U/ âˆ‚ p = (1/4) Â· (1)p2 6) Find âˆ‚ U/ âˆ‚ x, given U = 5x 1/2 y 1/3 . âˆ‚ U/ âˆ‚ x = (1/2)5x1/2 y 1/3 (treating "y" as a constant) 7) Find âˆ‚ U/ âˆ‚ y, given U = 2x + 3y + xy. âˆ‚ U/ âˆ‚ y = 0 + 3 + x (treating "x" as a constant) 8) Find âˆ‚ U/ âˆ‚ y, given U = (1+x)(1+y) (hint: remember the product rule of differentiation) âˆ‚ U/ âˆ‚ y = 0*(1+y) + (1+x)*(1) (note: in this partial derivative, treat "x" as a constant) âˆ‚ U/ âˆ‚ y = (1+x) 1...
View
Full Document
 Fall '08
 Dumas
 Economics, Microeconomics, Calculus, Derivative, Department of Economics and Finance, Dr. Chris Dumas, Calculus Review Example

Click to edit the document details