Download Solutions for Calculus Review Examples

UNC-Wilmington
Department of Economics and Finance
ECN 321
Dr. Chris Dumas
Solutions to Calculus Review Example Problems
1)
y = 23*ln(5x)
115 23
1
5 =

dy/dx = 23 
5x
5x
x
2)
y = 43x - 6x6 - 5*(3x - 7)3 (hint: use the "chain rule" on the (3x - 7)3 part)
dy/dx = 43 -36x5 - 5·3·(3x - 7)2·(3)
dy/dx = 43 -36x5 - 45·(3x - 7)2
3)
y = 5 - 6x - 4z2
y/x = -6
(note: in this partial derivative, treat "z" as a constant)
4)
Given y = 5 - 6x - 4p2 , 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 -(4p2)/6
x/y = -1/6
(note: in this partial derivative, treat "p" as a constant)
5)
Given 4 = -2U +7z4q - 1/(2p) , find U/p.
(hint: rearrange the equation for U, then take the derivative)
U = -4/2 +(7z4q)/2 - [1/(2p)]/2
U = -4/2 +(7z4q)/2 - (1/4)(1/p)
 1
  (note: in this partial derivative, treat "z" and "q" as constants)
U/p = -(1/4) ·
p  p 

U/p = -(1/4) · p 1 
p
U/p = -(1/4) · (-1)p-2
6)
Find U/x, given U = 5x1/2y1/3 .
U/x = (1/2)5x-1/2y1/3 (treating "y" as a constant)
7)
Find U/y, given U = 2x + 3y + xy.
U/y = 0 + 3 + x
(treating "x" as a constant)
8)
Find U/y, given U = (1+x)(1+y)
(hint: remember the product rule of differentiation)
U/y = 0*(1+y) + (1+x)*(1) (note: in this partial derivative, treat "x" as a constant)
U/y = (1+x)
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