FA06Ma.9Hw10solutions

# FA06Ma.9Hw10solutions - Math 1a Section 1 Solutions to...

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

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.

Unformatted text preview: Math 1a, Section 1 Solutions to Homework 10 1. 6.17. 21, 22, 31 p249 (a) Given f ( x ) = x x . Express as f ( x ) = exp ( x log x ). Therefore, f ( x ) = exp ( x log x )(log x + 1) (1) (b) Given f ( x ) = (1 + x )(1 + e x 2 ). The derivative is f ( x ) = (1 + e x 2 ) + (1 + x ) e x 2 2 x (2) (c) Given f ( x ) = (sin x ) cos x + (cos x ) sin x . Express f ( x ) = e cos x log sin x + e sin x log cos x (3) Then the derivative is given by f ( x ) = e cos x log sin x (- sin x log sin x + cos 2 x sin x ) + e sin x log cos x (cos x log cos x- sin 2 x cos x ) (4) 2. 6.26.2 p268 By the fundamental theorem of calculus, 2 f ( x ) f ( x ) = f ( x ) sin x 2 + cos x (5) Therefore, when f ( x ) 6 = 0, f ( x ) = sin x 2(2 + cos x ) (6) Therefore, f ( x ) = Z sin xdx 2(2 + cos x ) = Z- d cos x 2(2 + cos x ) =- 1 2 log(2 + cos x ) + C (7) Finally, note that f 2 (0) = 0, hence C = 1 2 log(3) 3. 6.26.13 Instead of the way it is in the book, we write p ( x ) = d + d 1 x + d 2 2 x 2 and f ( x ) = e x p ( x ). First, we differentiate)....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

FA06Ma.9Hw10solutions - Math 1a Section 1 Solutions to...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online