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1
**Rate of change / Differentiation (3)**

Differentiating Using differentiating

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**If y = xn = nxn-1 The Key Bit dy dx**

The general rule (very important) is :- If y = xn dy dx = nxn-1 E.g. if y = x2 = 2x dy dx E.g. if y = x3 = 3x2 dy dx E.g. if y = 5x4 = 5 x 4x3 = 20x3 dy dx

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**Find for these functions :- Gradient at x=-2**

Warm-up dy dx Find for these functions :- Gradient at x=-2 dy dx y= 2x2 y = 2x5 y = 5x2 + 10x + 5 y = x3 + x2 + x y = 2x4 - 4x2 + 7 = 4x = 10x4 = 10x + 10 = 3x2 + 2x +1 = 8x3 - 8x = -8 = 10x(-2)4=160 = = -10 = 3(-2)2+2(-2)+1 = = 9 = 8(-2)3 -8(-2) = 8 x-8 – 8 x-2 = = -48

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**A differentiating Problem**

The gradient of y = ax3 + 4x2 – 12x is 2 when x=1 What is a? dy dx = 3ax2 + 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 = 4x3 - ax2 + 10x is 6 when x=1 What is a?**

dy dx = 12x2 – 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 normals

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**Function Notation y = x3 – 12x f(x)= x3 – 12x f’(x)= 3x2 - 12**

We have seen: if dy dx = 3x2 - 12 Instead of ‘y’ we may use the function notation f(x) f(x)= x3 – 12x If dy dx then is replace by f’(x) f’(x)= 3x2 - 12 so f’(x) represent the differential/gradient function

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**Linear graphs y = 5x + 7 y = 2x - 1 y - intercept Gradient**

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

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**Definition y = 3x - 11.31 y = 3x + 2/3 y = 3x y = 3x + 8 y = 3x + 84**

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

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**Equations of Tangents y=x2 What is the equation of the tangent?**

The tangent to the curve is the gradient at that point (3,9) y x What is the equation of the tangent? y=x2 y = mx + c Substitute gradient: 9 = 6x3 + c c = = -9 y = 6x - 9 = 2x dy dx When x=3; = 2x3 = 6 dy dx

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**Equations of Tangents - try me**

y=3x2 + 4x + 1 - differentiate - gradient at x =1 - y = mx + c - find c (1,8) x What is the equation of the tangent? y=3x2 + 4x + 1 y = mx + c Substitute gradient: 8 = 10x1 + c c = = -2 y = 10x - 2 = 6x + 4 dy dx When x=1; = 6+4 = 10 dy dx

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Perpendicular Lines If two lines with gradients m1 and m2 are perpendicular, then:

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**Equations of Normals normal mT x mN = -1**

y=x2 The normal is always perpendicular to the tangent normal mT x mN = -1 y (3,9) x What is the equation of the normal? When x=3; = 2x3 = 6 dy dx y = mx + c Substitute gradient: 9 = -1/6 x 3 + c c = /6 = 9 + 1/2 y = -1/6 x + 9 1/2 mT x mN = -1 6 x mN = -1 mN = -1/6

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**Equations of Normals normal mT x mN = -1**

y=x2 The normal is always perpendicular to the tangent normal mT x mN = -1 y (3,9) x What is the equation of the normal? When x=3; = 2x3 = 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 mT x mN = -1 6 x mN = -1 mN = -1/6

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**Equations of Normal - try me**

y=3x2 + 4x + 1 - differentiate - gradient at x =1 - mt x mn = -1 - y = mnx + c - find c (1,8) normal x What is the equation of the normal? = 6x + 4 dy dx y = mnx + c Substitute gradient: 8 = -1/10 x 1 + c c = 8 + 1/10 = 8 1/10 y = -1/10 x + 8 1/10 When x=1; = 6+4 = 10 dy dx mT x mN = -1 10 x mN = -1 mN = -1/10

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1 Implicit Differentiation. 2 Introduction Consider an equation involving both x and y: This equation implicitly defines a function in x It could be defined.

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