c1 chapter 7 differentiation dr j frost ([email protected]) last modified: 13 th october...
TRANSCRIPT
Gradient of a curve
The gradient of the curve at a given point can be found by:1. Drawing the tangent at that point.2. Finding the gradient of that tangent.
How could we find the gradient?
δx represents a small change in x and δy represents a small change in y.
We want to find the gradient at the point A.
Suppose we’re finding the gradient of y = x2 at the point A(3, 9).
Gradient = 6 ?
How could we find the gradient?
How could we find the gradient?
For the curve y = x2, we find the gradient for these various points.Can you spot the pattern?
Point (1, 1) (4, 16) (2.5, 6.25) (10, 100)
Gradient 2 8 5 20
For y = x2, gradient = 2x?
Let’s prove it...
Proof that gradient of y = x2 is 2x
(x, x2)
(x + h, (x+h)2)
Suppose we add some tiny value, h, to x. Then:
The “lim” bit means “what this expression approaches as h tends towards 0”
The h disappears as h tends towards 0.
δx
δy?
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Further considerations(This slide is intended for Further Mathematicians only)
You may be wondering why we couldn’t just set h to be 0 immediately. Why did we have to expand out the brackets and simplify first?
If h was 0 at this stage, we’d have 0/0.This is known as an indeterminate form (i.e. it has no value!)
We don’t like indeterminate forms, and want to find some way to remove them. In this particular case, just expanding the numerator resolves the problem.
There are 7 indeterminate forms in total:0/0, 00, inf/inf, inf – inf, 1inf, inf^0, and 0 x inf
Notation
y = x2dydx
= 2x
f(x) = x2 f’(x) = 2x
Leibniz's notation
Lagrange’s notation
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These are ways in which we can express the gradient.
Your Turn: What is the gradient of y = x3?
(x, x3)
(x + h, (x+h)3)
δx
δy?
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Differentiating xn
Can you spot the pattern?
y x2 x3 x4 x5 x6
dy/dx 2x 3x2 4x3 5x4 6x5
If y = xn, then dy/dx = nxn-1! ?
If y = axn, then dy/dx = anxn-1
In general, scaling y also scales the gradient?
y = x7 dy/dx = 7x6 y = x10 dy/dx = 10x9
y = 2x3 dy/dx = 6x2 f(x) = 2x3 f’(x) = 6x2
f(x) = x2 + 5x4 f’(x) = 2x + 20x3 y = 3x1/2 dy/dx = 3/2 x-1/2
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Differentiating cx and c
x
y
y = 3
xWhat is the gradient of the line y = 3x?How could you show it using differentiation?
y = 3x = 3x1
Then dy/dx = 3x0 = 3
x
y = 4What is the gradient of the line y = 4?How could you show it using differentiation?
y = 4 = 4x0
Then dy/dx = 0x-1 = 0
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Differentiating cx and c
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Exercises
Function Gradient Point(s) of interest Gradient at this point
dy/dx = 10x4 (2, 64) 160
f’(x) = 21x2 (3, 189) 189
dy/dx = 2 + 2x (4, 24) 10
f’(x) = 3x2 – x-2 (2, 11.5) 11.75
f’(x) = 4x3 (2, 16) 32
dy/dx = 3x2 (3, 31), (-3, -23) 27
3
Be sure to use the correct notation for the gradient.
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Test your knowledge so far...
𝑑𝑦𝑑𝑥
=15𝑥2−8 𝑥13+2
𝑑2 𝑦𝑑𝑥2
=30 𝑥− 83𝑥− 23
Edexcel C1 May 2012
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Turning more complex expressions into polynomials
We know how to differentiate things in the form . Where possible, put expressions in this form.
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Exercise 7E (Page 115)
Use standard results to differentiate.
a) c) e) f) h) j) l)
Find the gradient of the curve with equation at the point A where:
c) and A is at d) and A is at
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2
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Finding equations of tangents
𝑥=3
Find the equation of the tangent to the curve when .
Function of gradient:
Gradient when :
-value when :
So equation of tangent:
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Finding equations of tangents
𝑥=3
Find the equation of the normal to the curve when .
Function of gradient:
Gradient when :
Perpendicular gradient when :
-value when :
So equation of normal:
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Second DerivativeWe can differentiate multiple times. For C1, you needn’t understand why we might want to do so.
Name Leibniz Notation Lagrange Notation
(Original expression/function)
First Derivative
Second Derivative
th Derivative
𝑦=5 𝑥3→ 𝑑2 𝑦𝑑 𝑥2
=30 𝑥
𝑦=→ 𝑑2𝑦𝑑𝑥2
=30 𝑥
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Equations of tangents and normalsEdexcel C1 Jan 2013
𝑑𝑦𝑑𝑥
=2−4 𝑥− 12
𝑦=−6𝑥+3
(9, -1)
Recap: If a line has gradient m and goes through , then it has equation:
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(questions on worksheet)
Equations of tangents and normalsEdexcel C1 Jan 2012
( 12 ,0)
(−8 , 178 )
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Equations of tangents and normals
Expanding gives Thus is as given.
So
We’re interested where the gradient is 20.
So (as before) or
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Equations of tangents and normalsEdexcel C1 Jan 2011
𝑑𝑦𝑑𝑥
=32𝑥2− 27
2𝑥12−8 𝑥−2
12
(4 )3−9 (4 )32+84+30=32−72+2+30=−8
Thus
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