Engineering Calculus Notes 251

Engineering Calculus Notes 251 - 239 3.3 DERIVATIVES A...

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Unformatted text preview: 239 3.3. DERIVATIVES A similar argument applied to the second term in parentheses (Exercise 6) yields f (x, y + △y ) − f (x, y ) = ∂f (x, y + δ2 ) △y ∂y where |δ2 | ≤ |△y | . This allows us to rewrite the quantity inside the absolute value in Equation (3.8) as ∂f ∂f (x, y ) △x + (x, y ) △y ∂x ∂y ∂f ∂f = f (x + △x, y + △y ) − f (x, y ) − (x, y ) △x + (x, y ) △y ∂x ∂y ∂f ∂f ∂f ∂f (x + δ1 , y + △y ) △x + (x, y + δ2 ) △y − (x, y ) △x + (x, y ) △y ∂x ∂y ∂x ∂y ∂f ∂f ∂f ∂f (x + δ1 , y + △y ) − (x, y ) △x + (x, y + δ2 ) − (x, y ) △y. = ∂x ∂x ∂y ∂y f (x + △x, y + △y ) − f (x, y ) + = Now, we want to show that this quantity, divided by (△x, △y ) = △x2 + △y 2 , goes to zero as (△x, △y ) → (0, 0). Clearly, |△x| △ x2 + △ y 2 |△y | △ x2 + △ y 2 ≤1 ≤ 1, so it suffices to show that each of the quantities in parentheses goes to zero. But as (△x, △y ) → 0, all of the quantities △x, △y , δ1 and δ2 go to zero, which means that all of the points at which we are evaluating the → partials are tending to − 0 = (x, y ); in particular, the difference inside each x pair of (large) parentheses is going to zero. Since each such quantity is being multiplied by a bounded quantity (△x/ △x2 + △y 2 , or △y/ △x2 + △y 2 ), the whole mess goes to zero. → This proves our assertion, that the affine function T− 0 f (− ) as defined by x → x → − ) at − = − . →→ x x0 Equation (3.7) has first-order contact with f ( x ...
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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.

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