{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW 7 - Tran Tony Homework 7 Due Oct 9 2007 3:00 am Inst...

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

Tran, Tony – Homework 7 – Due: Oct 9 2007, 3:00 am – Inst: Samuels 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points ±ind the derivative oF f when f ( x ) = 4 x cos 6 x. 1. f 0 ( x ) = 4 cos 6 x + 24 x sin 4 x 2. f 0 ( x ) = 24 cos 6 x - 4 x sin 6 x 3. f 0 ( x ) = 24 cos 6 x + 6 x sin 6 x 4. f 0 ( x ) = 4 cos 4 x - 4 x sin 6 x 5. f 0 ( x ) = 4 cos 6 x - 24 x sin 6 x correct Explanation: Using the Formulas For the derivatives oF sine and cosine together with the Chain Rule we see that f 0 ( x ) = (4 x ) 0 cos 6 x + 4 x (cos 6 x ) 0 = 4 cos 6 x - 24 x sin 6 x. keywords: derivative, trig Function, chain rule 002 (part 1 oF 1) 10 points ±ind the value oF f 0 (0) when f ( x ) = (1 - 2 x ) - 4 . Correct answer: 8 . Explanation: Using the chain rule and the Fact that ( x α ) 0 = α x α - 1 , we obtain f 0 ( x ) = 8 (1 - 2 x ) 5 . At x = 0, thereFore, f 0 (0) = 8 . keywords: derivative, chain rule 003 (part 1 oF 1) 10 points ±ind f 0 ( x ) when f ( x ) = p x 2 - 2 x . 1. f 0 ( x ) = 2( x - 1) p x 2 - 2 x 2. f 0 ( x ) = x - 1 x 2 - 2 x correct 3. f 0 ( x ) = ( x - 1) p x 2 - 2 x 4. f 0 ( x ) = 1 2 ( x - 1) p x 2 - 2 x 5. f 0 ( x ) = 2( x - 1) x 2 - 2 x 6. f 0 ( x ) = x - 1 2 x 2 - 2 x Explanation: By the Chain Rule, f 0 ( x ) = 1 2 x 2 - 2 x (2 x - 2) . Consequently, f 0 ( x ) = x - 1 x 2 - 2 x . keywords: derivative, square root, chain rule 004 (part 1 oF 1) 10 points

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

View Full Document
Tran, Tony – Homework 7 – Due: Oct 9 2007, 3:00 am – Inst: Samuels 2 Find f 0 ( x ) when f ( x ) = 1 4 x - x 2 . 1. f 0 ( x ) = x - 2 (4 x - x 2 ) 3 / 2 correct 2. f 0 ( x ) = 2 - x ( x 2 - 4 x ) 1 / 2 3. f 0 ( x ) = x - 2 (4 x - x 2 ) 1 / 2 4. f 0 ( x ) = x - 2 ( x 2 - 4 x ) 3 / 2 5. f 0 ( x ) = 2 - x ( x 2 - 4 x ) 3 / 2 6. f 0 ( x ) = 2 - x (4 x - x 2 ) 3 / 2 Explanation: By the Chain Rule, f 0 ( x ) = - 1 2(4 x - x 2 ) 3 / 2 (4 - 2 x ) . Consequently, f 0 ( x ) = x - 2 (4 x - x 2 ) 3 / 2 . keywords: derivative, square root, chain rule 005 (part 1 of 1) 10 points Determine f 0 ( x ) when f ( x ) = 1 - x 1 - 2 x 2 . 1. f 0 ( x ) = 1 + 2 x (1 - 2 x 2 ) 1 / 2 2. f 0 ( x ) = 2 x - 1 (1 - 2 x 2 ) 3 / 2 correct 3. f 0 ( x ) = 1 - 2 x (1 - 2 x 2 ) 1 / 2 4. f 0 ( x ) = 2 x - 1 (1 - 2 x 2 ) 1 / 2 5. f 0 ( x ) = 1 - 2 x (1 - 2 x 2 ) 3 / 2 6. f 0 ( x ) = 1 + 2 x (1 - 2 x 2 ) 3 / 2 Explanation: By the Product and Chain Rules, f 0 ( x ) = - 1 (1 - 2 x 2 ) 1 / 2 + 4 x (1 - x ) 2(1 - 2 x 2 ) 3 / 2 = - (1 - 2 x 2 ) + 2 x (1 - x ) (1 - 2 x 2 ) 3 / 2 . Consequently,
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

HW 7 - Tran Tony Homework 7 Due Oct 9 2007 3:00 am Inst...

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

View Full Document
Ask a homework question - tutors are online