Slope of a skateboard: 1·4 1·3 1·0 0·8 0·6 0 –1·9 –1·6 –0·2 –1·4 0·3 0·9 1·7 2 1 How steep is a hill?

Slide 3:

… the less accurate the slope The longer the skateboard ….

Slide 4:

and looking at the slope at one spot… P Zooming in on part of the hill

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P We get a more accurate measure … if we move the wheels closer together

Slide 6:

P We get a more accurate measure … if we move the wheels closer together

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P Definition of the slope of a hill at a point: The slope of a hill at point P is the LIMIT of the slopes of skateboards with one wheel at point P, as the other wheel approaches the wheel at P.

Differentiation from first principle :

Differentiation from first principle

Slide 9:

P y x (x, y) Graph y = f(x) y = f (x)

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P Now take a point a distance from P: x y (x + x, y + y) (x, y) y x y = f (x)

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y = f (x) P What is the slope of this line? y2 – y1 x2 – x1 y y x x

Slide 12:

P What happens as x gets smaller? x y

Slide 13:

P What happens as x gets smaller? x y

Slide 14:

y P What happens as x gets smaller? x

Slide 15:

x P What happens as x gets smaller? y

Slide 16:

x P What happens as x gets smaller? y

Slide 17:

P Slope becomes nearer to the slope of the tangent What happens as x gets smaller?

Slide 18:

P The slope of the tangent to the curve f ' (x)

Slide 19:

y + ∆y = x2 + 2x∆x + ∆x2 Differentiate x2 with respect to x from first principles. Let y = x2 Let x change to x + Dx, with a corresponding change in y to y + Dy Subtract Divide by Dx y + ∆y = (x + ∆x)2 = (x + ∆x)(x + ∆x) = x2 + x∆x + x∆x + ∆x2 y = x2 = 2x∆x + ∆x2 = 2x + ∆x ∆y Cancel 20

= 2x∆x + ∆x2 + x∆x = (x + ∆x)(x + ∆x) – 3(x + ∆x) + x∆x – 3∆x y + ∆y = x2 + 2x∆x + ∆x2 – 3x – 3∆x (b) Differentiate x2 – 3x with respect to x from first principles. Let y = x2 – 3x Let x change to x + Dx, with a corresponding change in y to y + Dy Subtract Divide by Dx y + ∆y = (x + ∆x)2 – 3(x + ∆x) = x2 + ∆x2 y = x2 – 3x = 2x + ∆x ∆y Cancel – 3x Each term inside the brackets changes sign 20 – 3∆x – 3

Slide 22:

y + ∆y = (x + ∆x)2 + 3(x + ∆x) y + ∆y = x2 + 2x∆x + ∆x2 + 3x + 3∆x (b) Differentiate x2 + 3x with respect to x from first principles. Let y = x2 + 3x Let x change to x + Dx, with a corresponding change in y to y + Dy Subtract Divide by Dx = (x + ∆x)(x + ∆x) + 3(x + ∆x) = x2 + x∆x + x∆x + ∆x2 + 3x + 3∆x y = x2 + 3x = 2x∆x + ∆x2 = 2x + ∆x ∆y + 3∆x Cancel + 3 20

Slide 23:

– ∆x + x∆x = 3(x + ∆x)(x + ∆x) – (x + ∆x) y + ∆y = 3x2 (b) Differentiate 3x2 – x from first principles with respect to x. Let y = 3x2 – x Let x change to x + Dx, with a corresponding change in y to y + Dy Subtract Divide by Dx y + ∆y = 3(x + ∆x) 2 – (x + ∆x) + ∆x2) y = 3x2 – x = 6x∆x – 3∆x2 = 6x – 3∆x ∆y Cancel = 3 (x2 + x∆x – x – ∆x Each term inside the brackets changes sign + 2x∆x – 1 1 20 + 6x∆x + 3∆x2 – x – ∆x

Differentiation from first principle :

Differentiation from first principle

Slide 25:

y = f (x) P f (x) x (x, f (x)) Graph y = f(x)

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y = f (x) P Now take a point a distance from P: f (x) x h (x, f (x)) f (x + h) (x + h, f (x + h))

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y = f (x) P What is the slope of this line? f (x) x h f (x + h) y2 – y1 x2 – x1

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x2 – x1 ( ) P What is the slope of this line? f (x) x h f (x + h) f (x + h) – f (x) y2 – y1 h + – x x What happens as h gets smaller?

Slide 29:

P What happens as h gets smaller? f (x) h f (x + h) f (x + h) – f (x) h x

Slide 30:

P What happens as h gets smaller? f (x) h f (x + h) f (x + h) – f (x) h x

Slide 31:

P What happens as h gets smaller? f (x) h f (x + h) f (x + h) – f (x) h x

Slide 32:

P What happens as h gets smaller? f (x) h f (x + h) f (x + h) – f (x) h x

Slide 33:

P What is the slope of this line? f (x) h f (x + h) f (x + h) – f (x) h x

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P What happens as h gets smaller? f (x) f (x + h) – f (x) h x Slope becomes nearer to the slope of the tangent

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P The slope of the tangent to the curve f (x) f (x + h) – f (x) h x f ' (x)

Slide 36:

f (x) = x 2
f (x + h) = (x + h)2 = x 2 + 2xh + h 2 = 2x + 0 = 2x Find the derivative of f(x) = x2

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