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Chapter 22
Differentiation
Starter
• What do you know about Gradient?
Objectives
• Define differentiation
• Find the gradient function
The branch of mathematics that deals with the finding and properties of
derivatives and integrals of functions
• The two main types are differential calculus and integral calculus.
Calculus
Differentiation, in mathematics, is the process of finding the
derivative, or rate of change, of a function.
In mathematics, a function is a relation between a set of inputs and a set
of permissible outputs with the property that each input is related to
exactly one output.
Can be written as when given in
the form y = ….
It means the derivative of y with
respect to x
https://teacher.desmos.com/activitybuilder/custom/5eb8586e18a18d3d6ae9dac7?collec
tions=6068b6deb21c6347a34cd9ef#preview/a1b3ccd1-715b-4a08-9d4d-931857b04d10
Derivative denotes by
Or
IGCSE Math Differentiation-introduction.pptx
Activity:
Test Your Understanding
𝑦 =𝑥
7
→
𝒅𝒚
𝒅𝒙
=𝟕 𝒙
𝟔
𝑦 =3 𝑥
10
→
𝒅𝒚
𝒅𝒙
=𝟑𝟎 𝒙
𝟗
𝑓 (𝑥 )=𝑥−10
=→
𝒅𝒚
𝒅𝒙
=−𝟏𝟎 𝒙−𝟏𝟏
𝑦 =5 𝑥
2
→
𝒅𝒚
𝒅𝒙
=𝟏𝟎 𝒙
𝟏
=𝟏𝟎 𝒙
?
?
?
?
0
?
1
2
3
4
5
Objective
• Differentiation of multiple term
Differentiating Multiple Terms
Differentiate
First thing to note:
If then
i.e. differentiate each term
individually in a sum/subtraction.
𝑑𝑦
𝑑𝑥
=2 𝑥 + 4
? ? ?
Therefore applying the usual rule:
Alternatively, if you compare to , it’s clear that the
gradient is fixed and .
Therefore applying the usual rule:
Alternatively, if you sketch , the line is horizontal, so the
gradient is 0.
Activity
𝑦 =2 𝑥
2
−3 𝑥 →
𝒅𝒚
𝒅𝒙
=𝟒 𝒙 −𝟑
1
2
3
4
5
𝑦 =4 − 9 𝑥
3
→
𝒅𝒚
𝒅𝒙
=−𝟐𝟕 𝒙
𝟐
𝑦 =5 𝑥+ 1→
𝒅𝒚
𝒅𝒙
=𝟓
𝑦 =𝑎𝑥 →
𝒅𝒚
𝒅𝒙
= 𝒂
𝑦 =6 𝑥 − 3+𝑝 𝑥
2
→
𝒅𝒚
𝒅𝒙
=𝟔+𝟐 𝒑𝒙
?
?
?
?
?
(where is a constant)
(where is a constant)
6
?
Objective
• Calculate Gradient of a curve at a point
Harder Example
Let
a) Find the gradient of at the point
b) Find the coordinates of the point on the graph of where the gradient is 8.
c) Find the gradient of at the points where the curve meets the line .
When Remember that the ‘gradient function’ allows you to find the
gradient for a particular value of .
Point is
This example is important!
Previously you used a value of to get the gradient . This time we’re doing the
opposite: using a known gradient to get the value of . We therefore substitute
for 8.
a
b
c First find point of intersection:
Solving, we obtain: or
When
When
Once you have your , you need to work out .
Ensure you use the correct equation!
?
?
?
Test Your Understanding
Let
a) Find the gradient of at the point
b) Find the coordinates of the point on the graph of where the gradient is 5.
c) Find the gradient of at the points where the curve meets the line .
When
Point is
a
b
c
Solving: or
When
When
?
?
?
Q y
dy
dx
Gradient of curve at x a

1
7 10
a 
2 2
1
3
2
x x
 4
a 
3 3
2 5 6
x x
  0
a 
4 4
2
3
x 1
a 
5 2
3 4 6
x x
   2
a 
6 3
2
5
x
 3
a 
7
17 3
x  7
a 
8 2
12 3
x x
 10
a 
9 2
5
4
x
x  1
a 
10 5 3 2
1 1
2 100
3 2
x x x
    1
a 
11 3
4x
1
2
a 
12 2
3x

2
3
a 
13
53 2
x 
4
7
a 
14 7 6 5 4
x x x x
   1
a 
y
dy
dx
x a

7 10
a 
2
1
3
2
x x

4
a 
3
2 5 6
x x
 
0
a 
4
2
3
x
1
a 2
3 4 6
x x
  
2
a 3
2
5
x

3
a 
17 3
x  7
a 
2
12 3
x x

10
a 
2
5
4
x
x 
1
a 5 3 2
1 1
2 100
3 2
x x x
   
1
a 
3
4x 1
2
a 
2
3x
 2
3
a 
53 2
x 4
7
a 
7 6 5 4
x x x x
  
1
a 
y
dy
dx
x a

7 10
a 
2
1
3
2
x x

4
a 
3
2 5 6
x x
 
0
a 
4
2
3
x
1
a 2
3 4 6
x x
  
2
a 3
2
5
x

3
a 
17 3
x  7
a 
2
12 3
x x

10
a 
2
5
4
x
x 
1
a 5 3 2
1 1
2 100
3 2
x x x
   
1
a 
3
4x 1
2
a 
2
3x
 2
3
a 
53 2
x 4
7
a 
7 6 5 4
x x x x
  
1
a 
Exercise 1
For each of the following, find the gradient function , and hence find the
gradient of the tangent to the curve when
The tangent to the curve has gradient 6. Determine the possible values
of .
For the curve , determine:
(a) The gradient of the tangent to the curve at
the point
(b) The point on the curve where the gradient is
5.
Find the points on the curve
where the gradient is 11.
1
2
3
4
a
b
c
d
e
f
g
?
?
?
?
?
?
?
?
?
?
?

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IGCSE Math Differentiation-introduction.pptx

  • 2. Starter • What do you know about Gradient?
  • 3. Objectives • Define differentiation • Find the gradient function
  • 4. The branch of mathematics that deals with the finding and properties of derivatives and integrals of functions • The two main types are differential calculus and integral calculus. Calculus
  • 5. Differentiation, in mathematics, is the process of finding the derivative, or rate of change, of a function. In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.
  • 6. Can be written as when given in the form y = …. It means the derivative of y with respect to x
  • 11. Test Your Understanding 𝑦 =𝑥 7 → 𝒅𝒚 𝒅𝒙 =𝟕 𝒙 𝟔 𝑦 =3 𝑥 10 → 𝒅𝒚 𝒅𝒙 =𝟑𝟎 𝒙 𝟗 𝑓 (𝑥 )=𝑥−10 =→ 𝒅𝒚 𝒅𝒙 =−𝟏𝟎 𝒙−𝟏𝟏 𝑦 =5 𝑥 2 → 𝒅𝒚 𝒅𝒙 =𝟏𝟎 𝒙 𝟏 =𝟏𝟎 𝒙 ? ? ? ? 0 ? 1 2 3 4 5
  • 13. Differentiating Multiple Terms Differentiate First thing to note: If then i.e. differentiate each term individually in a sum/subtraction. 𝑑𝑦 𝑑𝑥 =2 𝑥 + 4 ? ? ? Therefore applying the usual rule: Alternatively, if you compare to , it’s clear that the gradient is fixed and . Therefore applying the usual rule: Alternatively, if you sketch , the line is horizontal, so the gradient is 0.
  • 14. Activity 𝑦 =2 𝑥 2 −3 𝑥 → 𝒅𝒚 𝒅𝒙 =𝟒 𝒙 −𝟑 1 2 3 4 5 𝑦 =4 − 9 𝑥 3 → 𝒅𝒚 𝒅𝒙 =−𝟐𝟕 𝒙 𝟐 𝑦 =5 𝑥+ 1→ 𝒅𝒚 𝒅𝒙 =𝟓 𝑦 =𝑎𝑥 → 𝒅𝒚 𝒅𝒙 = 𝒂 𝑦 =6 𝑥 − 3+𝑝 𝑥 2 → 𝒅𝒚 𝒅𝒙 =𝟔+𝟐 𝒑𝒙 ? ? ? ? ? (where is a constant) (where is a constant) 6 ?
  • 15. Objective • Calculate Gradient of a curve at a point
  • 16. Harder Example Let a) Find the gradient of at the point b) Find the coordinates of the point on the graph of where the gradient is 8. c) Find the gradient of at the points where the curve meets the line . When Remember that the ‘gradient function’ allows you to find the gradient for a particular value of . Point is This example is important! Previously you used a value of to get the gradient . This time we’re doing the opposite: using a known gradient to get the value of . We therefore substitute for 8. a b c First find point of intersection: Solving, we obtain: or When When Once you have your , you need to work out . Ensure you use the correct equation! ? ? ?
  • 17. Test Your Understanding Let a) Find the gradient of at the point b) Find the coordinates of the point on the graph of where the gradient is 5. c) Find the gradient of at the points where the curve meets the line . When Point is a b c Solving: or When When ? ? ?
  • 18. Q y dy dx Gradient of curve at x a  1 7 10 a  2 2 1 3 2 x x  4 a  3 3 2 5 6 x x   0 a  4 4 2 3 x 1 a  5 2 3 4 6 x x    2 a  6 3 2 5 x  3 a  7 17 3 x  7 a  8 2 12 3 x x  10 a  9 2 5 4 x x  1 a  10 5 3 2 1 1 2 100 3 2 x x x     1 a  11 3 4x 1 2 a  12 2 3x  2 3 a  13 53 2 x  4 7 a  14 7 6 5 4 x x x x    1 a  y dy dx x a  7 10 a  2 1 3 2 x x  4 a  3 2 5 6 x x   0 a  4 2 3 x 1 a 2 3 4 6 x x    2 a 3 2 5 x  3 a  17 3 x  7 a  2 12 3 x x  10 a  2 5 4 x x  1 a 5 3 2 1 1 2 100 3 2 x x x     1 a  3 4x 1 2 a  2 3x  2 3 a  53 2 x 4 7 a  7 6 5 4 x x x x    1 a  y dy dx x a  7 10 a  2 1 3 2 x x  4 a  3 2 5 6 x x   0 a  4 2 3 x 1 a 2 3 4 6 x x    2 a 3 2 5 x  3 a  17 3 x  7 a  2 12 3 x x  10 a  2 5 4 x x  1 a 5 3 2 1 1 2 100 3 2 x x x     1 a  3 4x 1 2 a  2 3x  2 3 a  53 2 x 4 7 a  7 6 5 4 x x x x    1 a 
  • 19. Exercise 1 For each of the following, find the gradient function , and hence find the gradient of the tangent to the curve when The tangent to the curve has gradient 6. Determine the possible values of . For the curve , determine: (a) The gradient of the tangent to the curve at the point (b) The point on the curve where the gradient is 5. Find the points on the curve where the gradient is 11. 1 2 3 4 a b c d e f g ? ? ? ? ? ? ? ? ? ? ?