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Tangents and Differentiation

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Presentation on theme: "Tangents and Differentiation"β€” Presentation transcript:

1 Tangents and Differentiation
The gradient at any point on any curve is defined as the GRADIENT FUNCTION How do we find the gradient function of a curve?

2 Repeat for smaller values of π‘₯
Curves do not have a constant rate of change, i.e. the gradient is not constant. How do we calculate the rate of change at any point on a curve? This is defined to be the gradient of the tangent drawn at that point. 𝑓(π‘₯) 𝑓(π‘Ž) π‘Ž Repeat for smaller values of π‘₯ π‘₯ 𝑓 π‘₯ = π‘₯ 2 π‘Žπ‘›π‘‘ π‘Ž=1 x 2 π‘š 𝑠𝑒𝑐 3

3 This shows a sequence of points 𝑄 1 , 𝑄 2 …
This shows a sequence of points 𝑄 1 , 𝑄 2 …. getting closer and closer to the point P. What does this show? 𝑓 π‘₯ = π‘₯ 2 π‘Žπ‘›π‘‘ π‘Ž=1 x 2 1.5 1.1 1.001 π‘š 𝑠𝑒𝑐 3 2.5 2.1 2.001

4 𝑓 π‘₯ = π‘₯ 2 π‘Žπ‘›π‘‘ π‘Ž=1 As we choose values of π‘₯ getting closer to π‘Ž=1, we see that the lines from P to each of the Q’s get nearer and nearer to becoming a tangent at P. If we let the point Q approach the point P along the curve, then the gradient of the line should approach the gradient of the tangent at P, and hence the gradient of the curve. π‘₯ π’Ž 𝒔𝒆𝒄 2 3 1.5 2.5 1.1 2.1 1.001 2.001 The values of π’Ž 𝒔𝒆𝒄 appear to be getting closer to 2. In the case we write: lim π‘₯ β†’1 π’Ž 𝒔𝒆𝒄 =2

5 Algebraic Method Let Q be the point 1+β„Ž, 1+β„Ž 2
Gradient of PQ = (1+β„Ž) 2 βˆ’1 1+β„Ž βˆ’1 = 1+2β„Ž+ β„Ž 2 βˆ’1 β„Ž = β„Ž( 2+β„Ž) β„Ž = 2+β„Ž As Q approaches closer and closer to P, β„Žβ†’0 π‘ π‘œ, 2+β„Žβ†’2 This agrees with the previous method. This finds the gradient at one point of curve, but how can we find the gradient at any point?

6 Algebraic Method To find the gradient at 𝑃 π‘₯, π‘₯ 2 , take a point Q π‘₯+β„Ž, π‘₯+β„Ž 2 π‘₯+β„Ž, π‘₯+β„Ž 2 on the curve close to P. Gradient of PQ = (π‘₯+β„Ž) 2 βˆ’ π‘₯ 2 π‘₯+β„Ž βˆ’π‘₯ = π‘₯ 2 +2π‘₯β„Ž+ β„Ž 2 βˆ’ π‘₯ 2 β„Ž = β„Ž(2π‘₯+β„Ž) β„Ž = 2π‘₯+β„Ž As Q approaches closer and closer to P, β„Žβ†’0 π‘ π‘œ 2π‘₯+β„Žβ†’2π‘₯ At any point on 𝑦= π‘₯ 2 the gradient of the curve is 2π‘₯. So for 𝑦= π‘₯ 2 the gradient function is 2π‘₯ 𝑦= π‘₯ 2 (π‘₯+β„Ž) 2 (π‘₯+β„Ž) 2 βˆ’ π‘₯ 2 π‘₯ 2

7 Differentiation – First Principles
Differentiation is the process of finding the gradient function of a curve. For any curve 𝑦=𝑓(π‘₯), the gradient can be defined as follows: Let 𝑃(π‘₯,𝑦) be any point on the curve and Q π‘₯+𝛿π‘₯, 𝑦+𝛿𝑦 be a point on the curve near P. 𝛿π‘₯ and 𝛿𝑦 are the small changes in π‘₯ and 𝑦 between P and 𝑄. Gradient of P𝑄 = 𝑦+𝛿𝑦 βˆ’π‘¦ π‘₯+𝛿π‘₯ βˆ’ π‘₯ = 𝛿𝑦 𝛿π‘₯ Gradient at P = lim 𝛿π‘₯β†’0 𝛿𝑦 𝛿π‘₯ = 𝑑𝑦 𝑑π‘₯ ( 𝑑𝑦 𝑑π‘₯ is the notation used for the gradient function or derivative of a curve)

8 lim 𝛿π‘₯β†’0 𝛿𝑦 𝛿π‘₯ = 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim 𝛿π‘₯β†’0 𝑓 π‘₯+𝛿π‘₯ βˆ’π‘“(π‘₯) 𝛿π‘₯
Definition The derivative of a function 𝑓 with respect to a point π‘₯ is defined by: lim 𝛿π‘₯β†’0 𝛿𝑦 𝛿π‘₯ = 𝑑𝑦 𝑑π‘₯ = 𝑓′(π‘₯) = lim 𝛿π‘₯β†’0 𝑓 π‘₯+𝛿π‘₯ βˆ’π‘“(π‘₯) 𝛿π‘₯ Find the derivative of 𝑓 π‘₯ =π‘₯ 3 𝑓 β€² π‘₯ = lim 𝛿π‘₯β†’0 𝑓 π‘₯+𝛿π‘₯ βˆ’π‘“(π‘₯) 𝛿π‘₯ = lim 𝛿π‘₯β†’0 (π‘₯+𝛿π‘₯) 3 βˆ’ π‘₯ 3 𝛿π‘₯ = lim 𝛿π‘₯β†’0 π‘₯ 3 +3 π‘₯ 2 𝛿π‘₯ +3π‘₯ (𝛿π‘₯) 2 + (𝛿π‘₯) 3 βˆ’ π‘₯ 3 𝛿π‘₯ = lim 𝛿π‘₯β†’0 3 π‘₯ 2 +3π‘₯ 𝛿π‘₯ + (𝛿π‘₯) 2 = 3 π‘₯ 2 The derivative of 𝒙 πŸ‘ is πŸ‘ 𝒙 𝟐


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