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Find the gradient of the tangent to: a y = x2 at x = 2 b y = 8 x2 at the point (9, 8 81

Mathematics for the International Student: Mathematics SL | 3rd Edition | ISBN: 9781921972089 | Authors: Sandra Haese, Michael Haese, Robert Haese, Mark Humphries, Marjut Maenpaa ISBN: 9781921972089 446

Solution for problem 4 Chapter 15

Mathematics for the International Student: Mathematics SL | 3rd Edition

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Mathematics for the International Student: Mathematics SL | 3rd Edition | ISBN: 9781921972089 | Authors: Sandra Haese, Michael Haese, Robert Haese, Mark Humphries, Marjut Maenpaa

Mathematics for the International Student: Mathematics SL | 3rd Edition

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Problem 4

Find the gradient of the tangent to: a y = x2 at x = 2 b y = 8 x2 at the point (9, 8 81 ) c y = 2x2 3x + 7 at x = 1 d y = 2x2 5 x at the point (2, 3 2 ) e y = x2 4 x2 at the point (4, 3 4 ) f y = x3 4x 8 x2 at x = 1

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8.1 Integration by Parts ❑ ❑ ❑ ❑ ❑ Deals with: ∫ xe dx, ∫ xcos(2 x)dx∫, l| | d∫ , e c(x)dx , etc… ❑ ❑ ❑ ❑ Consider: u(x) & v(x) 2❑different❑able functions: (❑*v)’=(u*v’)+(u❑*v) &(integrate ' with respect to x) uv= ∫ uv )dx+∫(u' v)dx ∫ uv' dx=uv- ∫ u' vdx ❑ ❑ ❑ ❑ ❑ ❑ udv=uv− vdu ❑ ∫❑ x x xe dx u=x dv=e dx du=dx v=e x Ex. ❑ *Let dv be the easiest part of the integrand ¿ ∫❑ ❑ ❑ ❑ x x x x❑ x x ∫ xe dx=xe − e∫dx ∫ xe dx=xe −e +C ❑ ❑ ❑ ❑ xcos(2x)dxu=cos (2x du=−2sin 2x )dx 2 . ❑ dv=xdxv= x 2 x2 ❑ x2 x2 −2 ❑ 2 x2 ❑ 2 cos(2x)−∫ (2sin (x ))= cos(2x)−( ) ∫ x (in(2x )x= cos(2 ) ∫ x (sin2x)dx 2 ❑ 2 2 2 ❑ 2 ❑ This is not optimal. Instead haveu=xdu=dx . sin2 x) v= dv=cos(2 x)dx . 2 ❑ ❑ xsi(2x ) sin2x ) xsin2x ) 1 xsi(2x) cos2x ) 2 − ❑ 2 dx= 2 − 2❑ sin(2x)dx= 2 + 4 +C Shoetcut: Tabular Method Limi

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Chapter 15, Problem 4 is Solved
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Textbook: Mathematics for the International Student: Mathematics SL
Edition: 3
Author: Sandra Haese, Michael Haese, Robert Haese, Mark Humphries, Marjut Maenpaa
ISBN: 9781921972089

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Find the gradient of the tangent to: a y = x2 at x = 2 b y = 8 x2 at the point (9, 8 81