Higher Mathematics Unit 1.3
Differential Calculus
Contents
1.
1.
2.
3.
4.
5.
6.
7.
8.
Basic Ideas.
Differentiating x n .
Tangents to Curves.
Stationary Points/Curve Sketching.
Increasing/Decreasing Functions.
Sketching Graph of f (x ) .
Optimisation.
References/Resources.
Basic Ideas
It is highly important that pupils understand the basic ideas of differentiation.
The big idea to be grasped is:
gradient of tangent
rate of change of f (x)
f (x )
Gradient function
It is very important that pupils understand the equivalence of these three quantities.
With weaker pupils this idea will have to be gone over many times.
The following idea is also important:
Q
P
As “Q slides down the curve, getting closer to P”, the gradient of chord PQ approaches ever closer
to the gradient of the tangent at P.
Teachers should avoid doing an excessive number of numerical examples, for this will only
confuse many pupils.
First principles differentiation (interesting for more able pupils) may be left until S6; it is not
required for H Grade nowadays.
Many pupils will need much patient encouragement before the above ideas are known and
understood.
3
3
 x 1
Previous knowledge of fractions and indices is essential revision. Eg recognising
2x 2
2.
Differentiation: The Basic Rule
It is sufficient at this level to just state the rule.
y  xn

dy
 nx n 1
dx
and
f ( x)  x n

f ( x)  nx n 1
And n may be an integer or a rational number.
Some teachers may wish to introduce the two notations separately, while others may choose to
introduce them together.
d 2
( x  2 x)  2 x  2 can be introduced at a later stage.
dx
Some will need help when differentiating quotients and products.
Eg y =x (x + 2)(x + 3)
= x(x2 +5x +6)
y = x3+5 x2 +6x etc (stress the importance of correct layout)
Pupils should be confident differentiating basic examples before tackling the type below.
Working may be set out as follows, but able candidates will not need to show all these steps:
x2  x
2 x
x2
x
y 1  1
2x 2 2x 2
1 x2 1 x
y
1 
1
2 x2 2 x2
1 3 1 1
y  x2  x2
2
2
dy 1 3 12 1 1  12

x 
x
dx 2 2
22
dy 3 12 1 1
 x 
1
dx 4
4 x2
dy 3
1

x
dx 4
4 x
y
Be careful to prevent pupils committing the following type of “howler”:
y
x2  x
2 x
1
y  2 x 2 ( x 2  x)
Make them divide out first!
3.
Tangents to Curves
The process here is simple:
(i)
(ii)
(iii)
Use the equation for f ( x) to obtain the coords of the point of tangency.
Differentiate and use f (x ) to find the gradient of the tangent at the point.
Use y  b  m( x  a) to find the equation of the tangent line.
For example “ y  x 2 . Find the equation of the tangent to this curve at the point where x  3 .”
x  3  y  9 . Hence the point of tangency is (3,9).
dy
dy
 mtgt  6 .
 2x .
x 3
 6.
dx
dx
y  b  m( x  a )
y  9  6( x  3)
y  6x  9
At first consider simple functions such as x 2 , x 3 , x , and so on. This will be useful in reinforcing
the pupils’ knowledge of basic graphs and it will help their understanding of what is going on
when we are doing calculus.
Sketch graphs should be drawn in the early examples, again to illustrate what is actually going
on.
Regular revision (for example, in a few spare minutes at the end of a period) will consolidate what
has been learnt.
4.
Stationary Points/Curve Sketching
The most common type of question has been “Find the coords of the stationary points and
determine their nature algebraically.”
For example: “ y  2 x3  3x 2  12 x  6 . Find the coords of the stationary points and determine
their nature algebraically. Justify your answer.”
y  2 x 3  3x 2  12 x  6
dy
 6 x 2  6 x  12
dx
dy
0
dx
6 x 2  6 x  12  0
At a stationary point
important this is stated
6( x 2  x  2)  0
6( x  2)( x  1)  0
x  2 or x  1 ( x coords of st. pts.)
x  2  y  14
x  1  y  13
Hence the st. pts. are (2,-14) and (-1,13).
( encourage pupils to give coordinates)
Now do a table to determine the slope.
x
dy
dx
-1
+
0
2
-
0
+
Slope
Hence (-1,13) is a maximum T.P. and (2,-14) is a minimum T.P.
Some teachers will prefer to put more detail into the table; it is a matter of personal choice.
It is very important to stress that (for example) the equation 3 x 2  3 has two roots.
3x 2  3  0
Perhaps it is best to do
3( x 2  1)  0
3( x  1)( x  1)  0
x  1 or
x 1
Example
The diagram shows part of the graph of the curve
With equation y  2 x 3  7 x 2  4 x  4 .
Find the x-coordinate of the maximum
turning point.
Answer
 1
 2
 3
 4
 5
(5)
f'(x) = ………….
6x2-14x +4
6x2-14x +4 = 0
(3x – 1)(x – 2)
x = 1/3
Note:
a) When doing Stationary Points it is important to do examples concerned with finding the
maximum or minimum value of a function on a closed interval, for the extreme values of a
(continuous) function on a closed interval must occur either at a stationary point or at one of
the end points of the interval – needs to be stressed.
b) Worth revising how to find x and Y intercept.
5.
Increasing/Decreasing Functions
f (x ) increasing  f ( x)  0
f (x ) non-decreasing  f ( x)  0
At this level the approach should be
And similarly for “decreasing” and “non-increasing”.
Sketch graphs illustrating these ideas should be done on the board.
Example 1
For which values of x is f ( x)  2 x3  3x 2  12 x  6 increasing?
[This is the function whose stationary points were found and investigated in 4,above.]
The surest way to approach this is to find the x coords of the stationary points and determine the
nature of the stationary points, just as in 4, above.
Now f ( x)increa sin g  f ( x)  0 , which gives, from the table, x  2 or
x  1 .
There are other approaches, such as solving an appropriate quadratic inequation, but the above one
is “slow but sure”, and is therefore recommended.
Some pupils will need help describing an interval such as  3  x  5 .
Example 2
f ( x) 
x3
 x2  2x  3 .
3
Show that f is never-decreasing.
We must find f (x ) and show that f ( x )  0 for all real x.
To show that f ( x )  0 we may either complete the square or use calculus to find the minimum
value of f (x ) .
Completing the square is preferred here.
Working may be set out as follows:
f (x ) never-decreasing  f ( x)  0 for all real x.
f ( x)  x 2  2 x  2
x 2  2 x  2  ( x  b) 2  c
 x 2  2bx  b 2  c
Equating coefficients, we get
2b  2
b 1
b2  c  2
1 c  2
c 1
 f ( x)  ( x  1) 2  1
Now ( x  1) 2  0 for all real x (sq. of q real number)
 ( x  1)2  1  1
for all real x
and so f ( x) is never decreasing, since f ( x )  0 for all real x.
Example
a) Write x 2  10 x  27 in the form (x + b)2 + c.
(2)
1
b) Hence show that the function g ( x)  x 3  5 x 2  27 x  2
3
is always increasing.
(4)
Answer
 1
 2
 3
 4
 5
 6
6.
(x – 5)2……
(x – 5)2 + 2
g’(x) =
stated explicitly
2
x  10 x  27
(x – 5)2 + 2
g’(x) > 0 for all x and so g(x) is increasing.
Sketching the Graph of the Derived Function
Weaker pupils will need a great deal of practice at this topic.
f (x )
The key idea is:
Gradient of tangent
A good approach is to look first for stationary points.
Then do a table of f (x ) .
a) x coordinates of the turning points become the roots
b) from the table, the positive gradients are positive y coordinates, similarly the negative
gradients become negative coordinates.
c) Marks are awarded for roots and basic symmetrical curve.
Example
y
(-1,5)
f (x )
O
x
(3,-1)
The above sketch shows a sketch of the graph of a cubic function f (x) .
Draw a sketch of the graph of the derived function, f (x ) .
Now f ( x) cubic  f (x ) quadratic ( f(x) quadratic  f (x ) linear). Hence the graph of f (x )
is a parabola.
Note also that there are stationary points when x  1 and x  3 .
This gives
x

f (x ) +
-1
0
-
3
0
+
Thus we have
y
-1
O
f (x )
3
x
Do many examples such as this, and also consider examples where there is a horizontal point of
inflexion.
There has never been a question on the Inverse process, i.e. given the graph of f (x ) , draw a
possible graph of f ( x) , but this is a worthwhile topic for abler pupils.
Question
The graph of the cubic function y = f(x) is
Shown in the diagram. There are turning points
at (1, 1) and (3, 5).
Sketch the graph of y = f’(x).
Answer
 1
 2

7.
3
only 2 intercepts on the x-axis at 1 and 3
function is positive between the roots and
negative out with
a parabola (symmetrical about midpoint of
x- intercept), stated or implied by the accuracy
of the diagram.
Optimisation
This is a substantial topic, which is found to be difficult by many S5 pupils.
Therefore much practice is needed.
Most questions on this topic consist of:
(i)
(ii)
A first part (modelling), i.e. obtaining an equation to give a volume/surface area/cost, etc.
A second part, which requires us to find the maximum/minimum of the function found in
part (i).
Part (ii) is usually easier than part (i), so pupils must be encouraged to attempt part (ii), even if
they are unable to complete part (i) fully.
Some Points to Note
(a)
Always consider the range of possible values of x. For example, if x measures a length or
area we must have x>0, and in some cases we might have an interval such as o  x  6.
8.
(b)
If the interval to be considered is closed, e.g. 0  x  6 , then we must calculate the
function values at the end points, for the maximum or minimum values might occur at one
of the end points of the interval or at the turning points.
(c)
Always state “At a stationary point, f ( x )  0 ”: there is a mark for this!
(d)
Always do a check for max/min turning point by doing a table of values of f (x ) .
(e)
In the early stages of this topic, graphical calculators might give some useful insights to
the nature of these problems.
References/Resources
1.
2.
H.H.M. pages 85-119.
M.I.A. pp58 - 83