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matrix. To compute the righthand Q r tsQ i Y p# 0 0 # ` F ` U 9 # C % % 8# qe da 9 g c 0 g c aY W % % 1# hfedbC F # ` U ()& # H # ' # ` ! U 0 XY W V0 % % 8# C S8# U Q I $! # R % R 0 & S 0 % 8# HE R 0 F % 8# 8E R ! 0 & I Q
R R by and R R # R I 0 & 0 % 0 % 1# R 0 % 8# R ! T & Q 2 R # R 0 & 0 % 0 % 1# R 0 % 8# 6! R . This implies % 8# 0 9 F % 01# HG9 % 01# E 819 % 0 # 32 0 % 8# 76 5 ( )& # & % ! # ! % ' $" Q 1# 0 I P! % HG9 % 8# 0 9 % 8# % 1# D2 0 0 9 40 % DC % 1# BA@9 # 32 2 0 5 1. This is not a detailed solution sheet, but rather a collection of hints and summary answers intended to help you along in preparation for the exam. 2. As in several of the other problems, the trick here is to restate the problem in such a way that you can apply the (multivariable) chain rule. Define R R by Math 321 x cos x 3. Define which leads to the correct answer. we get Then is equivalent to , and therefore, by the chain rule, . Therefore, using practice problems for first prelim Some hints and solutions . R by ` where we view the lefthand side as a side of this identity, use and multiply and Then by and hence to work out the answer. Fall 2006 f w de C hw hw T hw 4. Implicit differentiation of the identity x x F1S8yw TH1x S8xS1xyw x x CF1S8yw gives x x F1S8yw TH1x F1S1yw x x Combining the formula for leads to the answer. which is often loosely written as with similar formul for and g 5. 6. This problem is very similar to homework problem B.5. See the online solution to problem B.5 for a hint. (a) (b) (c) n m A fk m l m )q m A m l m Tut n o n h ml o m fk m l m rsA fk m l m )q m A fk m l rq m m l k ji m n p n m o p o n m p o n o m k m l m rq m A fk k El Ajh o kn p o n m m k i
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 Fall '06
 SJAMAAR

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