{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

diff_tut_sol2

# diff_tut_sol2 - Differentiation II Solutions to...

This preview shows pages 1–3. Sign up to view the full content.

Differentiation II Solutions to Differentiation II Exercises. 1. f(x) = cos 2x f(0) = 1 f (x) = –2 sin 2x f (0) = 0 f (x) = –4 cos 2x f (0) = –2 2 f ″′ (x) = 8 sin 2x f ″′ (0) = 0 f (4) (x) = 16 cos 2x f (4) (0) = 2 4 ( 29 ( 29 ( 29 ( 29 ( 29 2 3 x x f x f 0 xf 0 f 0 f 0 ... 2! 3! ′′ ′′′ = + + + + 2 2 4 4 6 6 2 x 2 x 2 x cos2x 1 ... 2! 4! 6! = - + - + 2. ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 n n n 1 n 1 n 2 n 2 n 3 n 2 f x 1 x f (0) 1 1 f x n 1 x f (0) n 1 n f x n n 1 1 x f (0) n n 1 1 n n 1 f x n n 1 n 2 1 x f (0) n n 1 n 2 1 n n 1 n 2 - - - - - - = + = = = + = = ′′ ′′ = - + = - = - ′′′ ′′′ = - - + = - - = - - f ″′ (x) = n(n–1)(n–2)(1+x) n–3 f ″′ (0) = n(n–1)(n–2)(1 n–3 ) = n(n–1)(n–2) ( 29 ( 29 ( 29 ( 29 ( 29 2 3 x x f x f 0 xf 0 f 0 f 0 ... 2! 3! ′′ ′′′ = + + + + ( 29 ( 29 ( 29 ( 29 2 3 n n n 1 x n n 1 n 2 x 1 x 1 nx ... 2! 3! - - - + = + + + + 3. a) f(x) = cos 2 x f(0) = 1 f (x) = –2 cos x sin x f (0) = 0 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 (4) (4) (5) (5) (6) (6) f x cos x f 0 1 f x 2cos x sin x f 0 1 f x sin 2x f x 2cos2x f 0 2 f x 4sin 2x f 0 0 f x 8cos2x f 0 8 f x 16sin 2x f 0 0 f x 32cos2x f 0 32 = = =- = =- ′′ ′′ =- =- ′′′ ′′′ = = = = =- = =- =- ( 29 ( 29 ( 29 ( 29 ( 29 2 3 x x f x f 0 xf 0 f 0 f 0 ... 2! 3! ′′ ′′′ = + + + + 2 4 6 2 4 6 2 2 2x 8x 32x cos x 1 .... 2 24 720 x 2x cos x 1 x .... 3 45 = - + - + = - + - + 1

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

View Full Document
Differentiation II b) x xlna x xln 2 now a e 2 e = = ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 xln 2 0 xln 2 2 2 xln 2 3 3 xln 2 f x e f 0 e 1 f x ln 2e f 0 ln 2 f x ln 2 e f 0 ln 2 f x ln 2 e f 0 ln 2 = = = = = ′′ ′′ = = ′′′ ′′′ = = ( 29 ( 29 ( 29 ( 29 ( 29 2 3 x x f x f 0 xf 0 f 0 f 0 ... 2! 3! ′′ ′′′ = + + + + ( 29 ( 29 ( 29 2 3 4 x x ln 2 x ln 2 x ln 2 2 1 x ln 2 ... 2! 3! 4! = + + + + + 4. a) ( 29 ( 29 ( 29 ( 29 x x x x f x e f (1) e f x e f (1) e f x e f (1) e f x e f (1) e = = = = ′′ ′′ = = ′′′ ′′′ = = ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 3 0 0 0 0 0 0 0 x x x x f x f x x x f x f x f x ... 2! 3! - - ′′ ′′′ = + - + + + ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 3 x 2 3 x x 1 e x 1 e e e x 1 e ... 2! 3! x 1 x 1 e e 1 x 1 ... 2! 3! - - = + - + + + - - = + - + + + b) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 2 2 3 3 2 2 f x f 2 1 x 3 f x 2 1 x 2 f x 2 1 x f 2 3 2 2!
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