Ch 2 Rules of Differentiation Notes

# Ans 2 9 and 4 827 3 27 m 23 y 6x 2 4x 7 23 6x 2 4x 7

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ans: (2, 9) and 4 827 , 3 27 m = 23 y / = 6x 2 4x + 7 23 = 6x 2 4x + 7 0 = 6x 2 4x 16 = 2(3x 2 2x 8) = 2(3x + 4)(x 2) x = 2, 4 3 4. Find the equation of the tangents to y = 2 x x which are perpendicular to x y 19 = 0 ans: x + y 4 = 0 and x + y + 4 = 0 m = 1 / 2 2 1 y x 2 2 1 1 x 2 2 2 x x 2 = 1 x = 1 and 1 (1, 3) and ( 1, 3) 4c

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17 Implicit Differentiation Find the equation of the tangent to x 2 + y 2 = 25 at A(3,4). Method I (explicitly) y 2 = 25 x 2 2 2 1 2 y 25 x (top of circle) or y 25 x (bottom of circle) 1 1 / 2 / 2 2 2 1 2 1 1 y (25 x ) ( 2x) or y (25 x ) ( 2x) 2 2 / / 1 2 2 2 x x y (top of circle) or y (bottom of circle) 25 x 25 x Find the slope of the tangent at the point (3, 4) on top of circle, so use / 1 2 x 3 y 4 25 x Method II (implicitly) take the derivative with respect to x: d dx x 2 + y 2 = 25 2 2 2 2 / / 2 2 d d (x y ) (25) dx dx d d d (x ) (y ) (25) dx dx dx d d 2x (x) 2y (y) 0 chain rule dx dx 2x 2yy 0 x x x y (top) or (bottom) y 25 x 25 x Find the slope of the tangent at the point (3, 4). ans: 3 4 2. Given y 2 + xy + 7y = 13, find y / . 2yy / +[ y + xy / ] + 7y / = 0 y / (2y + x + 7) = y / y y 2y x 7 3. Find the equation of the tangent to 2y x x y at (6, 9) x 2 + xy = 2y 2x + y + xy / = 2y / 2x + y = y / (2 x) 5a -5 5 -5 5 0
18 / 2x y y 2 x at (6, 9) y / = 3 4 equation: 3x + 4y + 18 = 0 OR x 2 = y(2 x) 2 1 / 1 2 2 / 2 / 2 y x (2 x) y 2x(2 x) ( 1)(2 x) ( 1)(x ) y x(2 x) [2(2 x) x] x(4 x) y (2 x) at x = 6 / 12 3 y 16 4 4. Find the equation of the tangent line to the curve (x + y) 3 = x 3 + y 3 at ( 1, 1). 2 / 2 2 / 2 / 2 2 2 / / 2 2 2 2 2 2 / 2 2 3(x y) (1 y ) 3x 3y y 3(x y) y (3)(x y) 3x 3y y y [(3(x y) 3y ] 3x 3(x y) 3x 3(x y) y 3(x y) 3y OR 3 2 2 3 3 3 2 / 2 / 2 / 2 / / 2 2 2 / 2 x 3x y 3xy y x y 6xy 3x y 3y 3x(2yy ) 0 2xy x y y 2xyy 0 y (x 2xy) 2xy y 2xy y y x 2xy at ( 1, 1) y / = 1, equation: x + y = 0 5. If xy = 9, 2 9 dy y y or 9x x dx x depending on approach. Assignment pg. 107 # 1 b, d, e, h, 2 a, d, 3 a, c plus: 1. Find the equation of the tangent to x 2 + x 3 y 2 y 3 = 13 at (1, 2). ans: 2 2 / 2 3 2x 3x y y 3y 2x y , at (1, 2) y / = 7 8 , equation: 7x 8y 23 = 0 5b

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19 Higher Derivatives Uses acceleration If displacement is s = f(t) and velocity is v / ds f (t) dt then the acceleration, a, will be 2 // 2 dv d ds d s a f (t) dt dt dt dt concavity of graphs determining whether you have a minimum or maximum value Notation y // or f // (x) or f 2 (x) or 2 2 d y dx eg. 1. Given y = 3x 3 4x 2 7, find find y // . ans: 18x 8 2. Given 2 x y x 1 , find y // . / 1 2 2 / 2 / 2 // 2 2 3 // 3 // 3 2 2 2 // 3 y 2x(x 1) ( 1)(x 1) (x ) y x(x 1) [2(x 1) x] y x(x 2)(x 1) y (x 2)(x 1) x(x 1) ( 2)(x 1) (x)(x 2) y (x 1) [(x 2)(x 1) x(x 1) 2x(x 2)] y (x 1) [x 3x 2 x x 2x 4x] 2 y (x 1) 3. Given x 2 + y 2 = 25, find y // . 2x + 2yy / = 0 / 1 // 1 2 / // 1 2 1 // 1 2 3 2 2 2 2 // 3 2 2 3 3 3 y xy y y ( 1)y y ( x) y y ( 1)(y )( xy )( x) y y x y y x (y x ) 25 y y [ y x ] y y y 5c
20 4. Given x 2 y 2 = 1, find y // . / / 1 // 1 2 / 2 2 // 1 2 1 1 2 3 3 2 2 3 3 2x 2yy 0 y xy y y ( 1)y y (x) y x 1 y y ( 1)y (xy )(x) y x y y [y x ] y y 5. Given s = 4t 3 + 8t, find acceleration at t = 3. ans: 24(3) = 72 6. Given f(x) = 3 1 x , evaluate f // (2). ans: 3 // 3 3 3x(x 4) y 4 (1 x ) at x = 2, y // = 2 3 7. Differentiate with respect to x (chain rule) 5x 3 , (5x) 3 , 5y 3 , 5x 3 y 3 , (5xy) 3 , (5 + y) 3 , 5u 3 Assignment pg. 111 # 1 a, d, e, h, 2 a, c 7 b Review p112 1 12 as needed plus: 1. Given y 2 6x = 10, find y // at x = 1. ans: 9 9 or 64 64 2. Given y 2 4x 3 = 4, find 2 2 d y dx at y = 6. ans: 4 3 3. Given 2 x y (x 1) , find y // . ans: 4 2x 4 (x 1) 5d
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