```Gradients and Tangents
e.g. Find the gradient of the line joining the
points with coordinates ( 1, 1 ) and ( 3, 7 )
Solution:
m
y 2  y1
x 2  x1

m
71
31

6
3
2
difference in the
y-values
( 3, 7 )x
7-1=6
( 1, 1 )x
3-1=2
difference in the
x-values
The gradient of a straight line is given by
the difference in the
y - values
m
the difference in the x  values
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
This branch of Mathematics is called
Calculus
The Gradient at a point on a Curve
Definition: The gradient of a point on a curve equals
the gradient of the tangent at that point.
e.g.
y  x
2
12
(2, 4)x
3
Tangent at ( 2, 4)
The gradient of the tangent at (2, 4) is
m
So, the gradient of the curve at (2, 4) is 4
12
3
4
The gradient changes as we move along a curve
e.g.
y  x
m  6
2
y  x
2
m  4
y  x
2
m  2
The Rule
for Differentiation
y  x
2
m0
The Rule
for Differentiation
y  x
2
m2
y  x
2
m4
y  x
2
m6
Differentiation from first principles
400
360
320
P
280
f(x+h)
240
200
160
120
80
A
f(x)40
0
0
0.5
x1
1.5
2
2.5
f( x  h )  f( x )
h
3
x3.5
+h
4
Differentiation from first principles
400
360
320
280
240
200
P
f(x+h)
160
120
80
A
f(x)40
0
0
0.5
x1
1.5
2
2.5
f( x  h )  f( x )
h
x +3 h
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
P
120
f(x+h)
80
A
f(x)40
0
0
0.5
x1
1.5
2
2.5
x+
h
f( x  h )  f( x )
h
3
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
120
P
80
f(x+h)
A
f(x)40
0
0
0.5
x1
1.5
2
x+
h
2.5
f( x  h )  f( x )
h
3
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
120
80
f(x+h)
f(x)40
A
P
0
0
0.5
x1
x 1.5
+h
2
2.5
f( x  h )  f( x )
h
3
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
120
80
A
40
f(x)
0
0
0.5
x1
1.5
2
2.5
3
L im it f( x  h )  f( x )
Gradient of tangent at A =
h 0
h
3.5
4
Differentiation from first principles
f(x) = x2
f( x  h )  f( x )
( x  h)  x
2

2
x  2 xh  h  x
2

h
h

(2 x  h ) h
h
 2x  h
h
f '( x ) 
L im it f( x  h )  f( x )
h 0
h
2
 2x
2

2 xh  h
h
2
Differentiation from first principles
f(x) = x3
f( x  h )  f( x )
( x  h)  x
3

3
x  3 x h  3 xh  h  x
3

2
h
h

h
(3 x  3 xh  h ) h
2
2
 3 x  3 xh  h
2
h
f '( x ) 
L im it f( x  h )  f( x )
h 0
h
2
 3x
2
3
3
3 x h  3 xh  h
2

2
h
2
Generally with
L im it f( x  h )  f( x )
h 0
h
h is written as δx
And f(x+ δx)-f(x) is written as δy
Limit
x  0
x
dy

y
dx
Generally
f ( x )  kx
y  kx
n
f ' ( x )  nkx
n 1
dy
n
 nkx
n 1
dx
f ( x )  kx
y  kx
n
f ' ( x )  nkx
n 1
dy
dx
n
 nkx
n 1
e.g. Find the coordinates of the points on the curve
3
y  x  8 x  7 where the gradient equals 4
= 4
We need to be able
to find these points
using algebra
Exercises
Find the coordinates of the points on the curves with
1.
y  x  4x  3
2
Ans: (-3, -6)
3
2
2. y  x  3 x  21 x  20 where the gradient is 3
( Watch out for the common factor in the
Ans: (-2, 2) and (4, -88)
Increasing and Decreasing Functions
•
An increasing function is one whose
gradient is always greater than or equal to
zero.
dy
 0
dx
•
for all values of x
A decreasing function has a gradient that
is always negative or zero.
dy
dx
 0
for all values of x
e.g.1 Show that y  x 3  4 x is an increasing
function
dy
2
3
 3x  4
Solution: y  x  4 x 
dx
dy
is the sum of
dx • a positive number ( 3 )  a perfect
square ( which is positive or zero for
all values of x, and
• a positive number ( 4 )

dy
dx
 0 for all values of x
so, y  x 3  4 x is an increasing function
e.g.2 Show that y  1 x 3  3 x 2  9 x is an
3
increasing function.
Solution: y 
1
3
dy
dx
x  3x  9x 
3
2
dy
 x  6x  9
dx
 0
for all values of x
2
The graphs of the increasing functions
3
2
3
y  x  4 x and y  13 x  3 x  9 x
y  x  4x
3
and
y
1
3
x  3x  9x
3
2
are
Exercises
1. Show that y   x 3 is a decreasing function and
sketch its graph.
2. Show that y  1 x 3  2 x 2  5 x is an increasing
3
function and sketch its graph.
Solutions are on the next 2 slides.
Solutions
1. Show that y   x 3 is a decreasing function and
sketch its graph.
Solution:
dy
dx
  3 x . This is the product of a
2
square which is always  0 and a negative number,
dy
so
 0 for all x. Hence y   x 3 is a
dx
decreasing function.
y  x
3
Solutions
2. Show that
function
Solution:
y
dy
1
3
3
2
x  2 x  5 x is an increasing
 x  4x  5 .
dx
dy
2
Completing the square:
 ( x  2)  1
dx
which is the sum of a square which is  0
2
and a positive number. Hence y is an increasing
function.
y
1
3
x  2x  5x
3
2
The equation of a tangent
e.g. 1 Find the equation of the tangent at the point
(-1, 3) on the curve with equation
3
2
y  x  3x  2x  1
3
2
Solution:
y

x

3
x
 2 at
x  a1 point and the gradient of
dy the tangent
=equal
-5
2
at
that
point
are

 3x  6x  2
dx
 At x = 1
(-1, 3)
x
m  3(  1)  6(  1)  2
2
 5
y  mx  c 
y  5 x  c
(-1, 3) on line:  3   5 (  1 )  c

2c
So, the equation of the tangent is y   5 x  2
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