1. C1 Chapter 7 Differentiation
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 13th
October 2013
2. 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.
3. 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.
4. Suppose we’re finding the gradient of y = x2
at the point A(3, 9).
Gradient = 6 ?
How could we find the gradient?
5. 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...
6. 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
?
?
?
7. 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
8. Notation
y = x2
dy
dx
= 2x
f(x) = x2
f’(x) = 2x
Leibniz's notation
Lagrange’s notation
?
?
These are ways in which we can express the gradient.
9. Your Turn: What is the gradient of y = x3
?
(x, x3
)
(x + h, (x+h)3
)
δx
δy
?
?
?
10. 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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?
11. Differentiating cx and c
x
y
y
=
3
x
What 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 = 4
What is the gradient of the line y = 4?
How could you show it using differentiation?
y = 4 = 4x0
Then dy/dx = 0x-1
= 0
?
?
13. 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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14. Test your knowledge so far...
𝑑𝑦
𝑑𝑥
=15 𝑥2
−8 𝑥
1
3
+2
𝑑2
𝑦
𝑑 𝑥2
=30 𝑥 −
8
3
𝑥
−
2
3
Edexcel C1 May 2012
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15. 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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16. 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
1
2
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17. 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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18. 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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19. Second Derivative
We 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 𝑥
3
→
𝑑2
𝑦
𝑑 𝑥
2
=30 𝑥
𝑦 =→
𝑑2
𝑦
𝑑 𝑥
2
=30 𝑥
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?
20. Equations of tangents and normals
Edexcel C1 Jan 2013
𝑑𝑦
𝑑𝑥
=2 −4 𝑥
−
1
2
𝑦 =− 6 𝑥 +3
(9, -1)
Recap: If a line has gradient m and goes through , then it has equation:
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(questions on
worksheet)