PracticeMidterm2Solutions

PracticeMidterm2Solutions - Practice Midterm 2 Solutions 1....

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Practice Midterm 2 Solutions 1. a) Using the chain rule dy dx = e x 3 +1 · 3 x 2 b) Using the change of base formula, y = ln cos x ln x so from the quotient and chain rules, dy dx = ln x · 1 cos x ( - sin x ) - 1 x ln cos x (ln x ) 2 c) Applying d dx to each side and using the product and chain rules 2 y dy dx + 6 x dy dx + 6 y + 2 x = 0 dy dx = - 3 y - x y + 3 x 2. Let be the distance from the origin to the point q . By the Pythagorean theorem, 2 = x 2 + y 2 . Applying d dt to each side and using the chain rule gives 2 d‘ dt = 2 x dx dt + 2 y dy dt d‘ dt = x · dx dt + y · dy dt Since y = x 2 , by the chain rule dy dt = 2 x dx dt , so d‘ dt = ( x + 2 x 3 ) x 2 + x 4 · dx dt = sign ( x ) 1 + 2 x 1 + x 2 · dx dt 3. Approximating sin by its tangent line at 0, sin 1 = sin π 180 sin 0 + (cos 0) ± π 180 - 0 ² = π 180 4. a) Method 1 Recognizing the limit as a derivative, lim b 2 b 2010 - 2 2010 b - 2 = d db ( b 2010 ) ³ ³ ³ ³ b =2 = 2010 · 2 2009
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This note was uploaded on 02/05/2011 for the course MATH 1A taught by Professor Wilkening during the Spring '08 term at Berkeley.

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PracticeMidterm2Solutions - Practice Midterm 2 Solutions 1....

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