{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# u-max-demo - OPMT 5701 Calculus for Utility Problems Kevin...

This preview shows pages 1–3. Sign up to view the full content.

OPMT 5701: Calculus for Utility Problems Kevin Wainwright October 14, 2003 1 Using Calculus For Utility Maximization Prob- lems 1.1 Review of Some Derivative Rules 1. Partial Derivative Rules: U = xy U/ x = y U/ y = x U = x a y b U/ x = ax a 1 y b U/ y = bx a y b 1 U = x a y b = x a y b U/ x = ax a 1 y b U/ y = bx a y b 1 U = ax + by U/ x = a U/ y = b U = ax 1 / 2 + by 1 / 2 U/ x = a ³ 1 2 ´ x 1 / 2 U/ y = b ³ 1 2 ´ y 1 / 2 2. Logarithm (Natural log) ln x (a) Rules of natural log If Then y = AB ln y = ln( AB ) = ln A + ln B y = A/B ln y = ln A ln B y = A b ln y = ln( A b ) = b ln A NOTE: ln( A + B ) 6 = ln A + ln B (b) derivatives IF THEN y = ln x dy dx = 1 x y = ln ( f ( x )) dy dx = 1 f ( x ) · f 0 ( x ) (c) Examples If Then y = ln( x 2 2 x ) dy/dx = 1 ( x 2 2 x ) (2 x 2) y = ln( x 1 / 2 ) = 1 2 ln x dy/dx = ³ 1 2 ´ ³ 1 x ´ = 1 2 x 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3. The Number e if y = e x then dy dx = e x if y = e f ( x ) then dy dx = e f ( x ) · f 0 ( x ) (a) Examples y = e 3 x dy dx = e 3 x (3) y = e 7 x 3 dy dx = e 7 x 3 (21 x 2 ) y = e rt dy dt = re rt 1.2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}