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12 - ELEMENTARY MATHEMATICS W W L CHEN and X T DUONG c W W...

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ELEMENTARY MATHEMATICS W W L CHEN and X T DUONG c ° W W L Chen, X T Duong and Macquarie University, 1999. This work is available free, in the hope that it will be useful. Any part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, with or without permission from the authors. Chapter 12 FURTHER TECHNIQUES OF DIFFERENTIATION 12.1. The Chain Rule We begin by re-examining a few examples discussed in the previous chapter. Example 12.1.1. Recall Examples 11.2.5 and 11.2.10, that for the function y = h ( x )=( x 2 +2 x ) 2 , we have d y d x = h 0 ( x )=4 x 3 +12 x 2 +8 x. On the other hand, we can build a chain and describe the function y = h ( x ) by writing y = g ( u )= u 2 and u = f ( x x 2 x. Note that d y d u =2 u and d u d x x , so that d y d u × d u d x u (2 x +2)=2( x 2 x )(2 x +2)=4 x 3 x 2 x. This chapter was written at Macquarie University in 1999.

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12–2 W W L Chen and X T Duong : Elementary Mathematics Example 12.1.2. Recall Example 11.3.3, that for the function y = h ( x ) = sin 2 x, we have d y d x = h 0 ( x ) = 2sin x cos x. On the other hand, we can build a chain and describe the function y = h ( x ) by writing y = g ( u )= u 2 and u = f ( x ) = sin x. Note that d y d u =2 u and d u d x = cos x, so that d y d u × d u d x u cos x = 2sin x cos x. Example 12.1.3. Recall Example 11.3.4, that for the function y = h ( x ) = sin2 x, we have d y d x = h 0 ( x ) = 2cos2 x. On the other hand, we can build a chain and describe the function y = h ( x ) by writing y = g ( u ) = sin u and u = f ( x )=2 x. Note that d y d u = cos u and d u d x , so that d y d u × d u d x = 2cos u = 2cos2 x. In these three examples, we consider functions of the form y = h ( x ) which can be described in a chain by y = g ( u ) and u = f ( x ), where u is some intermediate variable. Suppose that x 0 ,x 1 R . Write u 0 = f ( x 0 ) and u 1 = f ( x 1 ), and write y 0 = g ( u 0 ) and y 1 = g ( u 1 ). Then clearly h ( x 0 g ( f ( x 0 )) and h ( x 1 g ( f ( x 1 )). Heuristically, we have h ( x 1 ) h ( x 0 ) x 1 x 0 = y 1 y 0 x 1 x 0 = y 1 y 0 u 1 u 0 × u 1 u 0 x 1 x 0 = g ( u 1 ) g ( u 0 ) u 1 u 0 × f ( x 1 ) f ( x 0 ) x 1 x 0 . If x 1 is close to x 0 , then we expect that u 1 is close to u 0 , and so the product g ( u 1 ) g ( u 0 ) u 1 u 0 × f ( x 1 ) f ( x 0 ) x 1 x 0 is close to g 0 ( u 0 ) f 0 ( x 0 ), while the product h ( x 1 ) h ( x 0 ) x 1 x 0 is close to h 0 ( x 0 ). It is therefore not unreasonable to expect the following result, although a formal proof is somewhat more complicated.
Chapter 12 : Further Techniques of Di±erentiation 12–3 CHAIN RULE. Suppose that y = g ( u ) and u = f ( x ) . Then d y d x = d y d u × d u d x , provided that the two derivatives on the right hand side exist. We can interpret the rule in the following way. As we vary x , the value u = f ( x ) changes at the rate of d u/ d x . This change in the value of u = f ( x ) in turn causes a change in the value of y = g ( u )at the rate of d y/ d u .

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12 - ELEMENTARY MATHEMATICS W W L CHEN and X T DUONG c W W...

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