Section 3.9-solutions - ) = 8 x sec 2 (4 x 2 ) ....

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to (aqt73) – Section 3.9 – isaacson – (55826) 1 This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind the linearization oF f ( x ) = 1 6 + x at x = 0. 1. L ( x ) = 1 6 p 1 + 1 12 x P 2. L ( x ) = 1 6 + 1 6 x 3. L ( x ) = 1 6 p 1 + 1 12 x P 4. L ( x ) = 1 6 - 1 6 x 5. L ( x ) = 1 6 p 1 - 1 6 x P 6. L ( x ) = 1 6 p 1 - 1 12 x P correct Explanation: The linearization oF f is the Function L ( x ) = f (0) + f (0) x . But For the Function f ( x ) = 1 6 + x = (6 + x ) 1 / 2 , the Chain Rule ensures that f ( x ) = - 1 2 (6 + x ) 3 / 2 . Consequently, f (0) = 1 6 , f (0) = - 1 12 6 , and so L ( x ) = 1 6 p 1 - 1 12 x P . 002 10.0 points ±ind the di²erential, dy , oF y = f ( x ) = tan(4 x 2 ) . 1. dy = 4 sec 2 (4 x 2 ) tan(4 x 2 ) + dx 2. dy = 4 sec 2 (4 x 2 ) tan(4 x 2 ) 3. dy = 4 sec 2 (4 x 2 ) tan(4 x 2 ) dx 4. dy = 8 x sec 2 (4 x 2 ) 5. dy = 8 x sec 2 (4 x ) + dx 6. dy = 8 x sec 2 (4 x 2 ) dx correct Explanation: The di²erential, dy , oF y = f ( x ) is given by dy = f ( x ) dx . On the other hand, d dx tan x = sec 2 x . Thus by the Chain Rule, f ( x
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Unformatted text preview: ) = 8 x sec 2 (4 x 2 ) . Consequently, dy = 8 x sec 2 (4 x 2 ) dx . keywords: dierential, trig Function, tan Func-tion, Chain Rule, 003 10.0 points to (aqt73) Section 3.9 isaacson (55826) 2 Find the diferential dy when y = 1 + sin x 3-sin x . 1. dy = 3 sin x (3-sin x ) 2 dx 2. dy =-cos x 3-sin x dx 3. dy = sin x (3-sin x ) 2 dx 4. dy =-4 cos x 3-sin x dx 5. dy = 4 cos x (3-sin x ) 2 dx correct 6. dy =-4 cos x (3-sin x ) 2 dx Explanation: Ater diferentiation o y = 1 + sin x 3-sin x using the quotient rule we see that dy = (3-sin x ) cos x + cos x (1 + sin x ) (3-sin x ) 2 dx . Consequently, dy = 4 cos x (3-sin x ) 2 dx ....
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Section 3.9-solutions - ) = 8 x sec 2 (4 x 2 ) ....

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