The presentation includes all the necessary material for review of Chapter 6.1 The Rule for Differentiation for A-level Pure Mathematics Cambridge International Examinations. It will be useful for students as the review materials and also for teachers as the teaching tips and presentations in the class.
The presentation was made by Oleksii Khlobystin (Alex) - Mathematics Teacher at Depu Foreign Language School, Chongqing City, China.
Personal website: www.visualcv.com/o-khlobystin
Personal email: o-khlobystin@yandex.comPresentation POWER POINT Chapter 6.1 The Rule for Differentiation, A-level Pure Mathematics CIE 9709
Made by Oleksii Khlobystin (Alex) - Mathematics Teacher at Depu Foreign Language School – Cambridge International Centre, Chongqing City, China.
Personal website: www.visualcv.com/o-khlobystin
Personal email: o-khlobystin@yandex.com
The Gradient of a Straight
given by
Line 1212xxyymwhere and are points on the line
),(11yx),(22yx
The gradient of a straight line is
The Gradient of a Straight Line
the points with coordinates and)1,1()7,3(
1212xxyym3261317m)7,3(x)1,1(x
difference in the
y-values
difference in
the x-
values
7 1 = 6
3 1 = 2
e.g. Find the gradient of the line joining
Solution:
difference in the x-valuesdifference in the y-values
values the in difference the values- the in difference thexym
We use this idea to get the gradient at a
point on a curve
Gradients are important as they measure the
rate of change of one variable with another.
For the graphs in this section, the gradient
measures how y changes with x
The gradient of a straight line is
given by
This branch of Mathematics is called
Calculus
Curve
The Gradient at a point on a
Definition: The gradient of a point on a curve
equals the gradient of the tangent at that
point.
e.g.
),(42Tangent
4312mSo, the gradient of the curve at (2, 4)
The gradient of the tangent at (2, 4) is
is 4
y
2x
12
(2, 4)x
3
at
The Gradient at a point on a Curve
a curve
e.g.
y 6mThe gradient changes as we move along
2x
For the curve we have the following
gradients: 2xyPoint on the
Gradient)4,2()9,3()1,1()0,0()1,1()4,2()9,3( 6420246At every point, the gradient is twice the x-
curve
value
For the curve we have the following
gradients: 2xyGradient6420246)4,()9,()1,()0,0()1,()4,()9,(211233At every point, the gradient is twice the x-
Point on the
curve
value
twice the x-value2xyThis rule can be written asxdxdy2The notation comes from the idea of the
gradient of a line being the ifference in the -valuesthe ifference in the - valuesd
At every point on the gradient is
d
y
x
twice the x-value2xyThis rule can be written asxdxdy2The notation comes from the idea of the
the ifference in the -valuesthe ifference in the - values
is read as “ dy by dx ”dxdyThe function giving the gradient of a curve is
At every point on the gradient is
called the gradient function
gradient of a line being
d
d
y
x
functions
23xdxdy3xy
34xdxdy4xy45xdxdy5xyThe rule for the gradient function of a
curve of the form nxy1nnxdxdyis
gradients at points on the curve3xyOther curves and their gradient
Although this rule won’t be proved, we can
illustrate it for by sketching the
“power to the front and
“subtract 1 from the power”
multiply”
Other curves and their gradient functions
y3xy
3xyGradient
dxdy)27,3(x
of
)12,2(x)27,3(x dxdyy
3xy
3xyGradient
of
dxdyy
3xy
3xyGradient
of )27,3(x)12,2(x)3,1(x
3xyy
3xyGradient
of )27,3(x dxdy)12,2(x)3,1(x)0,0(x
3xyy
3xyGradient
dxdy)27,3(x)12,2(x)3,1(x)0,0(x )3,1(x
of
3xyy
3xyGradient
dxdy)27,3(x)12,2(x)3,1(x)0,0(x )3,1(x )12,2(x
of
3xyy
3xyGradient
dxdy)27,3(x)12,2(x)3,1(x)0,0(x )3,1(x )12,2(x )27,3(x
of
y
23xdxdy3xy
dxdy)27,3(x)12,2(x)3,1(x)0,0(x )3,1(x )12,2(x )27,3(x
know their gradients xy1y
1yxyThe gradients of the functions and
can also be found by the rule but as
they represent straight lines we already
gradient =
gradient =
0
1
Functions:
34x23x45xx21dxdy3x4x5x2xxy10Summary of Gradient
Summary of Gradient Functions:
Point
e.g.1 Find the gradient of the curve at
The Gradient Function and Gradient at a
the point (2, 12). 3xydxdyAt x = 2, the
n: 3xy23x12
gradient 2)2(3dxdymSolutio
The Gradient Function and Gradient at a Point
3xy8121,23xdxdySolution
43221213,mxAt
the point where x 1 4xyAt x = 1, m = 434xdxdySolution
Exercise
point
:
:
1. Find the gradient of the curve at
s
2. Find the gradient of the curve at the
Exercises
The process of finding the gradient
where a is a constant. naxy
The rule for differentiating can be extended
to curves of the form
function is called differentiation.
The gradient function is called the
derivative.
by 2 3xy
3xytangent at x =
More Gradient
Functions
e.g. Multiplying by 2 multiplies the gradient
gradient =
3
1
tangent at x = 1 More Gradient Functions
by 2 3xy3xytangent at x =
32xygradient =
tangent at x =
1
6
e.g. Multiplying by 2 multiplies the gradient
gradient =
3
1
tangent at x = 1 tangent at x = 1
constantnxThe rule can also be used for sums and
32xy 232xdxdy26xMultiplying by a constant,
multiplies the gradient by that
differences of terms.
bx c
cx d
e.g.
2+
bx
ax
3
+2
2
For y
3
ax
dy
dx
For y
n
1
i
i
a x
i
c
dy
dx
n
1
i
a i x
i
i
1
e.g. 3752123xxxydxdy 710232xx2321xx257
on the curve 42323xxxySolution
function: 1492xxdxdyWhen x = 1, gradient m
1)1(4)1(9212mUsing Gradient
e.g. Find the gradient at the point where x = 1
=
Functions
: Differentiating to find the gradient
Using Gradient Functions
The gradient at a point on a curve is
defined as the gradient of the tangent at
that point
The function that gives the gradient of a
curve at any point is called the gradient
function
SUMMARY
1nanxdxdynaxyis
The process of finding the gradient
function is called differentiating
• “power to the front and
• “subtract 1 from the
multiply”
power”
The rule for differentiating terms of the
form
SUMMARY
following curves:
1. 375223xxxy at the
point )1,1(
n 3,1mx4,1mxWhe
71062xxdxdy26122xxdxdy2xdxdyFind the gradients at the given points on the
point )5,1(
2. 423423xxxy at the
point )1,2(Whe
3. 12221xxy at the
n 0,2mxWhe
Exercise
s
n
Exercises
4. )4)(2(xxy at )9,1(
rule )822xxy 22xy22xdxdy xdxdy2at )2,2(xxxxy23 0,1mxWhe
xxxy235.
n42mxWhe
( Multiply out the brackets before using the
rule )
(Divide out before using the
Exercise
s
Find the gradients at the given points on the
following curves:
n
Exercises
Chapter 6.1 The Rule for Differentiation-Alex.ppt
