Engineering Calculus Notes 274

20 since x y x y to complete the proof we need

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x ∂f /∂y = (△x, △y ) |ε| |∂f /∂y | |△x| ≤ |ε| △y 1+ |∂f /∂y | △x (3.20) (since (△x, △y ) ≤ |△x| + |△y |). To complete the proof, we need to find △y an upper bound for 1 + △x on the right side. To this end, we come back to Equation (3.19), this time moving just the second term to the left, and then take absolute values, using the triangle inequality (and (△x, △y ) ≤ |△x| + |△y |): |△y | ∂f ∂f ≤ |△x| + |ε| |△x| + |ε| |△y | . ∂y ∂x Gathering the terms involving △x on the left and those involving △y on the right, we can write |△y | ∂f − |ε| ∂y ≤ |△x| ∂f + |ε| ∂x or, dividing by the term on the left, |△y | ≤ |△x| |∂f /∂x| + |ε| |∂f /∂y | − |ε| . (3.21)...
View Full Document

This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

Ask a homework question - tutors are online