# HW15 - Jolley Garrett Homework 15 Due Dec 4 2007 3:00 am...

This preview shows pages 1–4. Sign up to view the full content.

Jolley, Garrett – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: R Heitmann 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Determine if lim x → ∞ tan - 1 1 + 2 x 2 5 + 4 x exists, and if it does, find its value. 1. limit = π 6 2. limit = 0 3. limit = π 3 4. limit does not exist 5. limit = π 4 6. limit = π 2 correct Explanation: Since 1 + 2 x 2 5 + 4 x -→ as x → ∞ , we see that lim x → ∞ tan - 1 1 + 2 x 2 5 + 4 x exists, and that the limit = tan - 1 = π 2 . keywords: limit, limit at infinity, arctan, 002 (part 1 of 1) 10 points Find the derivative of f ( x ) = 2 sin - 1 ( e 3 x ) . 1. f 0 ( x ) = 6 e 3 x 1 + e 6 x 2. f 0 ( x ) = 2 e 3 x 1 - e 6 x 3. f 0 ( x ) = 2 1 + e 6 x 4. f 0 ( x ) = 2 e 3 x 1 + e 6 x 5. f 0 ( x ) = 6 1 - e 6 x 6. f 0 ( x ) = 6 1 + e 6 x 7. f 0 ( x ) = 6 e 3 x 1 - e 6 x correct 8. f 0 ( x ) = 2 1 - e 6 x Explanation: Since d dx sin - 1 x = 1 1 - x 2 , d dx e ax = ae ax , the Chain Rule ensures that f 0 ( x ) = 6 e 3 x 1 - e 6 x . keywords: 003 (part 1 of 1) 10 points Determine f 0 ( x ) when f ( x ) = tan - 1 x 5 - x 2 · . ( Hint : first simplify f .) 1. f 0 ( x ) = 1 5 - x 2 correct 2. f 0 ( x ) = 5 5 - x 2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Jolley, Garrett – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: R Heitmann 2 3. f 0 ( x ) = 5 5 + x 2 4. f 0 ( x ) = x x 2 + 5 5. f 0 ( x ) = x x 2 - 5 Explanation: If tan θ = x 5 - x 2 , then by Pythagoras’ theorem applied to the right triangle p 5 - x 2 x θ 5 we see that sin θ = x 5 . Thus f ( x ) = sin - 1 x 5 · . Consequently, f 0 ( x ) = 1 5 - x 2 . Alternatively, we can differentiate f using the Chain Rule and the fact that d dx tan - 1 x = 1 1 + x 2 . . keywords: 004 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 5 tan - 1 ( e 2 x ) + 3 e - 2 x . 1. f 0 ( x ) = 1 1 - e 4 x ( 2 e 2 x + 3 e - 2 x ) 2. f 0 ( x ) = 1 1 + e 4 x ( 2 e 2 x + 3 e - 2 x ) 3. f 0 ( x ) = 2 1 + e 4 x ( 2 e 2 x + 3 e - 2 x ) 4. f 0 ( x ) = 2 1 - e 4 x ( 2 e 2 x + 3 e - 2 x ) 5. f 0 ( x ) = 2 1 + e 4 x ( 2 e 2 x - 3 e - 2 x ) cor- rect 6. f 0 ( x ) = 1 1 + e 4 x ( 2 e 2 x - 3 e - 2 x ) Explanation: By the Chain Rule, f 0 ( x ) = 2 5 e 2 x 1 + e 4 x - 3 e - 2 x since d dx tan - 1 x = 1 1 + x 2 ( e 2 x ) 2 = e 4 x . The expression for f 0 can now be simplified by bringing the right hand side to a common denominator, for then f 0 ( x ) = 2 5 e 2 x - 3 e - 2 x (1 + e 4 x ) 1 + e 4 x = 2 5 e 2 x - 3 e - 2 x - 3 e 2 x 1 + e 4 x . Consequently, f 0 ( x ) = 2 1 + e 4 x ( 2 e 2 x - 3 e - 2 x ) . keywords: 005 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = ˆ tan - 1 x 2 - ln r 2 + x 2 - x ! .
Jolley, Garrett – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: R Heitmann 3 1. f 0 ( x ) = - 4 x 2 16 - x 4 correct 2. f 0 ( x ) = 2 x 2 16 + x 4 3. f 0 ( x ) = - 4 x 2 16 + x 4 4. f 0 ( x ) = 6 16 - x 4 5. f 0 ( x ) = 2 x 2 16 - x 4 6. f 0 ( x ) = 6 16 + x 4 Explanation: Since ln r 2 + x 2 - x = 1 2 h ln(2 + x ) - ln(2 - x ) i , we see that d dx ˆ ln r 2 + x 2 - x ! = 1 2 1 2 + x + 1 2 - x = 2 4 - x 2 . On the other hand, d dx tan - 1 x 2 · = 1 2 ˆ 1 1 + x 2 4 ! = 2 4 + x 2 . Thus f 0 ( x ) = 2 4 + x 2 - 2 4 - x 2 = 2(4 - x 2 ) - 2(4 + x 2 ) (4 + x 2 )(4 - x 2 ) .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern