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Rate of change / Differentiation (3) Differentiating Differentiating Using differentiating Using differentiating

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The Key Bit The general rule (very important) is :- If y = x n dy dx = nx n-1 E.g. if y = x 2 = 2x dy dx E.g. if y = x 3 = 3x 2 dy dx E.g. if y = 5x 4 = 5 x 4x 3 = 20x 3 dy dx dy dx

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Find for these functions :- dy dx y= 2x 2 y = 2x 5 y = 5x 2 + 10x + 5 y = x 3 + x 2 + x y = 2x 4 - 4x 2 + 7 dy dx dy dx dy dx dy dx dy dx = 4x = 10x 4 = 10x + 10 = 3x 2 + 2x +1 = 8x 3 - 8x Gradient at x=-2 = -8 = 10 x (-2) 4 =160 = -20 + 10 = -10 = 3(-2) 2 +2(-2)+1 = 12 -4 +1 = 9 = 8(-2) 3 -8(-2) = 8 x -8 – 8 x -2 = -64 +16 = -48 Warm-up

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A differentiating Problem The gradient of y = ax 3 + 4x 2 – 12x is 2 when x=1 What is a? dy dx = 3ax 2 + 8x -12 When x=1 dy dx = 3a + 8 – 12 = 2 3a - 4 = 2 3a = 6 a = 2

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Try this The gradient of y = 4x 3 - ax 2 + 10x is 6 when x=1 What is a? dy dx = 12x 2 – 2ax + 10 When x=1 dy dx = 12 -2a +10 = 6 22 - 2a = 6 16 = 2a a = 8

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Rate of change / Differentiation (3 pt2) Equations of tangents Equations of tangents Equations of normals Equations of normals

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Function Notation f(x)= x 3 – 12x dy dx = 3x 2 - 12 Instead of ‘y’ we may use the function notation f(x) dy dx is replace by f’(x) If then so f’(x)= 3x 2 - 12 y = x 3 – 12x We have seen: if f’(x) represent the differential/gradient function

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Linear graphs Gradient y - intercept m and c will always be numbers in your examples y = 5x + 7 y = 2x - 1 Increase in y Increase in x Gradient = dy dx

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Definition Parallel lines are ones with the same slope/gradient. i.e. the number in front of the ‘x’ is the same y = 3x + 8 y = 3x - 1.5 y = 3x + 2 / 3 y = 3x + 84 y = 3x - 21 y = 3x + 43 y = 3x + 1 y = 3x - 3 y = 3x - 11.31 y = 3x

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y x Equations of Tangents y=x 2 The tangent to the curve is the gradient at that point (3,9) What is the equation of the tangent? y=x 2 = 2x dy dx When x=3; = 2 x 3 = 6 dy dx y = mx + c Substitute gradient: 9 = 6 x 3 + c c = 9 - 18 = -9 y = 6x - 9

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y x Equations of Tangents - try me y=3x 2 + 4x + 1 (1,8) What is the equation of the tangent? y=3x 2 + 4x + 1 = 6x + 4 dy dx When x=1; = 6+4 = 10 dy dx y = mx + c Substitute gradient: 8 = 10 x 1 + c c = 8 - 10 = -2 y = 10x - 2 - differentiate - gradient at x =1 - y = mx + c - find c

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Perpendicular Lines If two lines with gradients m 1 and m 2 are perpendicular, then:

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y x Equations of Normals y=x 2 (3,9) What is the equation of the normal? When x=3; = 2 x 3 = 6 dy dx y = mx + c Substitute gradient: 9 = - 1 / 6 x 3 + c c = 9 - - 3 / 6 = 9 + 1 / 2 y = - 1 / 6 x + 9 1 / 2 m T x m N = -1 6 x m N = -1 m N = -1/6 The normal is always perpendicular to the tangent normal m T x m N = -1

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y x Equations of Normals y=x 2 (3,9) What is the equation of the normal? When x=3; = 2 x 3 = 6 dy dx y = mx + c Substitute gradient: 9 = - 1 / 6 x 3 + c c = 9 + 3 / 6 = 9 + 1 / 2 y = - 1 / 6 x + 9 1 / 2 m T x m N = -1 6 x m N = -1 m N = -1/6 The normal is always perpendicular to the tangent normal m T x m N = -1

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y x Equations of Normal - try me y=3x 2 + 4x + 1 (1,8) What is the equation of the normal? = 6x + 4 dy dx When x=1; = 6+4 = 10 dy dx y = m n x + c - differentiate - gradient at x =1 - m t x m n = -1 - y = m n x + c - find c normal m T x m N = -1 10 x m N = -1m N = - 1 / 10 Substitute gradient: 8 = - 1 / 10 x 1 + c c = 8 + 1 / 10 = 8 1 / 10 y = - 1 / 10 x + 8 1 / 10

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