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Calculus Review Example Solutions - 4q/2[1(2p/2 U =-4/2(7z...

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UNC-Wilmington ECN 321 Department of Economics and Finance Dr. Chris Dumas Solutions to Calculus Review Example Problems 1) y = 23*ln(5x) dy/dx = 5 x 5 1 23 = x 23 x 5 115 = 2) y = 43x - 6x 6 - 5*(3x - 7) 3 (hint: use the "chain rule" on the (3x - 7) 3 part) dy/dx = 43 -36x 5 - 5·3·(3x - 7) 2 ·(3) dy/dx = 43 -36x 5 - 45·(3x - 7) 2 3) y = 5 - 6x - 4z 2 y/ x = -6 (note: in this partial derivative, treat "z" as a constant) 4) Given y = 5 - 6x - 4p 2 , find x/ y . (hint: rearrange the equation to get x alone on the left-hand side of the equation, then take the derivative) x = 5/6 -y/6 -(4p 2 )/6 x/ y = -1/6 (note: in this partial derivative, treat "p" as a constant) 5) Given 4 = -2U +7z 4q - 1/(2p) , find U/ p. (hint: rearrange the equation for U, then take the derivative) U = -4/2 +(7z
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Unformatted text preview: 4q )/2 - [1/(2p)]/2 U = -4/2 +(7z 4q )/2 - (1/4)(1/p) βˆ‚ U/ βˆ‚ p = -(1/4) Β· βˆ‚ βˆ‚ p 1 p (note: in this partial derivative, treat "z" and "q" as constants) βˆ‚ U/ βˆ‚ p = -(1/4) Β· ( 29 1 p p-βˆ‚ βˆ‚ βˆ‚ U/ βˆ‚ p = -(1/4) Β· (-1)p-2 6) Find βˆ‚ U/ βˆ‚ x, given U = 5x 1/2 y 1/3 . βˆ‚ U/ βˆ‚ x = (1/2)5x-1/2 y 1/3 (treating "y" as a constant) 7) Find βˆ‚ U/ βˆ‚ y, given U = 2x + 3y + xy. βˆ‚ U/ βˆ‚ y = 0 + 3 + x (treating "x" as a constant) 8) Find βˆ‚ U/ βˆ‚ y, given U = (1+x)(1+y) (hint: remember the product rule of differentiation) βˆ‚ U/ βˆ‚ y = 0*(1+y) + (1+x)*(1) (note: in this partial derivative, treat "x" as a constant) βˆ‚ U/ βˆ‚ y = (1+x) 1...
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