HW15-solutions

# HW15-solutions - white(taw933 – HW15 – ben-zvi...

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Unformatted text preview: white (taw933) – HW15 – ben-zvi – (55600) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the derivative of f ( x ) = 2 sin − 1 ( x/ 4) . 1. f ′ ( x ) = 4 √ 1- x 2 2. f ′ ( x ) = 2 √ 1- x 2 3. f ′ ( x ) = 2 √ 16- x 2 correct 4. f ′ ( x ) = 4 √ 16- x 2 5. f ′ ( x ) = 8 √ 16- x 2 6. f ′ ( x ) = 8 √ 1- x 2 Explanation: Use of d dx sin − 1 ( x ) = 1 √ 1- x 2 , together with the Chain Rule shows that f ′ ( x ) = 2 radicalbig 1- ( x/ 4) 2 parenleftBig 1 4 parenrightBig . Consequently, f ′ ( x ) = 2 √ 16- x 2 . 002 10.0 points Find the derivative of f ( x ) = parenleftBig tan − 1 (2 x ) parenrightBig 2 . 1. f ′ ( x ) = 2 4 + x 2 tan − 1 (2 x ) 2. f ′ ( x ) = sec 2 (2 x ) tan(2 x ) 3. f ′ ( x ) = 2 1 + 4 x 2 tan − 1 (2 x ) 4. f ′ ( x ) = 4 4 + x 2 tan − 1 (2 x ) 5. f ′ ( x ) = 4 sec 2 (2 x ) tan(2 x ) 6. f ′ ( x ) = 4 1 + 4 x 2 tan − 1 (2 x ) correct Explanation: Since d dx tan − 1 x = 1 1 + x 2 , the Chain Rule gives d dx tan − 1 (2 x ) = 2 1 + 4 x 2 . Using the Chain Rule yet again, therefore, we see that f ′ ( x ) = 4 1 + 4 x 2 tan − 1 (2 x ) . keywords: derivative, inverse tan, Chain Rule, 003 10.0 points Determine f ′ ( x ) when f ( x ) = tan − 1 parenleftBig x √ 3- x 2 parenrightBig . ( Hint : first simplify f .) 1. f ′ ( x ) = √ 3 √ 3- x 2 2. f ′ ( x ) = 1 √ 3- x 2 correct 3. f ′ ( x ) = x √ x 2- 3 white (taw933) – HW15 – ben-zvi – (55600) 2 4. f ′ ( x ) = x x 2 + 3 5. f ′ ( x ) = √ 3 √ 3 + x 2 Explanation: If tan θ = x √ 3- x 2 , then by Pythagoras’ theorem applied to the right triangle radicalbig 3- x 2 x θ √ 3 we see that sin θ = x √ 3 . Thus f ( x ) = sin − 1 parenleftBig x √ 3 parenrightBig . Consequently, f ′ ( x ) = 1 √ 3- x 2 . Alternatively, we can differentiate f using the Chain Rule and the fact that d dx tan − 1 x = 1 1 + x 2 . . 004 10.0 points Find the derivative of f when f ( x ) = 5 tan − 1 x- 2 ln radicalbigg 1 + x 1- x . 1. f ′ ( x ) = 7 + 3 x 2 1- x 4 2. f ′ ( x ) = 3- 7 x 2 1- x 4 correct 3. f ′ ( x ) = 3- 7 x 2 1- x 2 4. f ′ ( x ) = 7- 3 x 2 1- x 2 5. f ′ ( x ) = 3 + 7 x 2 1- x 4 6. f ′ ( x ) = 7- 3 x 2 1- x 4 Explanation: Since ln radicalbigg 1 + x 1- x = 1 2 [ln(1 + x )- ln(1- x )] , we see that d dx bracketleftBigg ln radicalbigg 1 + x 1- x bracketrightBigg = 1 2 bracketleftbigg 1 1 + x + 1 1- x bracketrightbigg = 1 1- x 2 . On the other hand, d dx bracketleftbig tan − 1 x bracketrightbig = 1 1 + x 2 . Thus f ′ ( x ) = 5 1 + x 2- 2 1- x 2 = 5(1- x 2 )- 2(1 + x 2 ) (1 + x 2 )(1- x 2 ) ....
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HW15-solutions - white(taw933 – HW15 – ben-zvi...

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