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C H A P T E R
22Differentiation
Techniques
ObjectivesTo differentiate functions having negative integer powers.
To understand and use the chain rule.
To differentiate rational powers.
To find second derivatives of functions.
To use limits to define continuity at a point.
To use differentiation techniques to sketch graphs of functions.
To apply differentiation techniques to solving problems.
The work in this chapter is not part of the content for VCE Mathematical Methods (CAS)
Units 1 & 2. It is included as a useful foundation for VCE Mathematical Methods (CAS)
Units 3 & 4.
22.1 Differentiating xn where n isa negative integerIn earlier chapters the differentiation of polynomial functions has been introduced and
applications of differentiation of such functions have been considered. In this section we add to
the family of functions for which we can find the derived functions. In particular, we will
consider functions which involve linear combinations of powers of x where a power may be a
negative integer.
e.g. f : R \ {0} → R, f (x) = x−1
f : R \ {0} → R, f (x) = 2x + x−1
f : R \ {0} → R, f (x) = x + 3 + x−2
Note: We have reintroduced the function notation which emphasises the need for consideration
of domain.
582
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Chapter 22 — Differentiation Techniques 583
Example 1
Let f : R \ {0} → R, f (x) = 1
x. Find f ′(x) by first principles.
Solution
y
x
x, P 1x
Q x + h, ––––1x + h
0
Gradient of chord PQ = f (x + h) − f (x)
x + h − x
=1
x + h− 1
xh
= x − (x + h)
(x + h)x× 1
h
= −h
(x + h)x× 1
h
= −1
(x + h)x
The gradient of the curve at P = limh→0
−1
(x + h)x= −1
x2= x−2, i.e. f ′(x) = −x−2
Example 2
Let f : R \ {0} → R, f (x) = x−3. Find f ′(x) by first principles.
Solution
y
x
(x, x–3)
(x + h, (x + h)–3)
0
Q
P
Gradient of chord PQ = (x + h)−3 − x−3
h
= x3 − (x + h)3
(x + h)3x3× 1
h
= x3 − (x3 + 3x2h + 3xh2 + h3)
(x + h)3x3× 1
h
= −3x2h − 3xh2 − h3
(x + h)3x3× 1
h
= −3x2 − 3xh − h2
(x + h)3x3
The gradient of the curve at P = limh→0
−3x2 − 3xh − h2
(x + h)3x3
= −3x2
x6
= −3
x4
= −3x−4
i.e. f ′(x) = −3x−4
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584 Essential Mathematical Methods 1 & 2 CAS
We are now in a position to state the generalisation of the result we found in Chapter 19.
For f (x) = xn , f ′(x) = nxn−1, n is a non-zero integer.
For f (x) = c, f ′(x) = 0, where c is a constant.
We note that for n ≤ −1 we take the domain of f to be R \ {0}, and for n ≥ 1 we take
the domain of f to be R.
Example 3
Find the derivative of x4 − 2x−3 + x−1 + 2, x �= 0.
Solution
If f (x) = x4 − 2x−3 + x−1 + 2, x �= 0
f ′(x) = 4x3 − 2(−3x−4) + (−x−2) + 2(0)
= 4x3 + 6x−4 − x−2, x �= 0
Example 4
Find the derivative f ′ of f : R \ {0} → R, f (x) = 3x2 − 6x−2 + 1.
Solution
f ′: R \ {0} → R, f ′(x) = 3(2x) − 6(−2x−3) + 1(0)
= 6x + 12x−3
Example 5
Find the gradient of the curve determined by the function
f : R \ {0}, f (x) = x2 + 1
xat the point (1, 2).
Solution
f ′: R \ {0} → R, f ′(x) = 2x + (−x−2)
= 2x − x−2
and f ′(1) = 2 − 1
= 1
The gradient of the curve is 1 at the point (1, 2).
Example 6
Show that the gradient of the function f : R \ {0} → R, f (x) = x−3 is always negative.
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Chapter 22 — Differentiation Techniques 585
Solution
f ′: R \ {0} → R, f ′(x) = −3x−4
= −3
x4
x−4 is positive for all x and thus f ′(x) < 0 for all x �= 0.
Exercise 22A
1 Differentiate each of the following with respect to x :Examples 3, 4
a 3x−2 + 5x−1 + 6 b3
x2+ 5x2 c
5
x3+ 4
x2+ 1
d 3x2 + 5
3x−4 + 2 e 6x−2 + 3x f
3x2 + 2
x
2 Find the derivative of each of the following:
a3z2 + 2z + 4
z2, z �= 0 b
3 + z
z3, z �= 0 c
2z2 + 3z
4z, z �= 0
d 9z2 + 4z + 6z−3, z �= 0 e 9 − z−2, z �= 0 f5z − 3z2
5z, z �= 0
3 a Carefully sketch the graph of f (x) = 1
x2, x �= 0.Examples 1, 2
b Let P be the point (1, f (1)) and Q the point (1 + h, f (1 + h)). Find the gradient of the
chord PQ.
c Hence find the gradient of the curve f (x) = 1
x2at x = 1.
d Find the equation of the normal to the curve with equation y = x−2 at the point (1, 1).
4 Find the gradient of each of the following curves at the given point:Example 5
a y = x−2 + x3, x �= 0, at(2, 8 1
4
)b y = x − 2
x, n �= 0, at
(4, 1
2
)
c y = x−2 − 1
x, x �= 0, at (1, 0) d y = x(x−1 + x2 − x−3), x �= 0, at (1, 1).
5 Find the x-coordinate of the point on the curve with equation f (x) = x−2 and gradient:
a 16 b −16
22.2 The chain ruleAn expression such as f (x) = (x3 + 1)2 may be differentiated by expanding and then
differentiating each term. This method is a great deal more tiresome for an expression
such as f (x) = (x3 + 1)30. We transform f (x) = (x3 + 1)2 into two simpler functions
defined by:
h(x) = x3 + 1(= u) and g(u) = u2(= y)
which are ‘chained’ together: xh−→u
g−→y
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586 Essential Mathematical Methods 1 & 2 CAS
We may exploit this connection to differentiate. We use Leibniz notation to explore
this idea.
Consider f (x) = (x3 + 1)2 at the point where x = 2.
u y
u
Q
R
S
u = x3 + 1
y = u2
x 00 91 2
δx
δyδu
δu
P
80
60
40
20(0, 1)
–2 –1 0 1 2
y
x
Z
W
δy
δx
y = (x3 + 1)2
When x = 2, u = 9 When u = 9, y = 81 When x = 2, y = 81
P is the point (2, 9) R is the point (9, 81) Z is the point (2, 81)
A section of a spreadsheet (below) illustrates the connection between the gradients of the
chords PQ, RS and WZ. It can be seen that as �x gets smaller so does �u and�y
�x= �y
�u× �u
�x.
By considering �u becoming smaller and hence �x becoming smaller, it is seen that
dy
dx= dy
du· du
dx
This is called the chain rule for differentiation.
x u y �x �u �y�u
�x
�y
�u
�u
�x× �y
�u
�y
�x
1.60000 5.09600 25.96922 0.40000 3.90400 55.03078 9.76000 14.09600 137.57696 137.57696
1.80000 6.83200 46.67622 0.20000 2.16800 34.32378 10.84000 15.83200 171.61888 171.61888
1.90000 7.85900 61.76388 0.10000 1.14100 19.23612 11.41000 16.85900 192.36119 192.36119
1.99000 8.88060 78.86504 0.01000 0.11940 2.13496 11.94010 17.88060 213.49614 213.49614
1.99900 8.98801 80.78425 0.00100 0.01199 0.21575 11.99400 17.98801 215.74816 215.74816
1.99990 8.99880 80.97840 0.00010 0.00120 0.02160 11.99940 17.99880 215.97480 215.97480
1.99999 8.99988 80.99784 0.00001 0.00012 0.00216 11.99994 17.99988 215.99748 215.99748
From the spreadsheet it can be seen that the gradient of u = x3 + 1 at x = 2 is 12, and the
gradient of y = u2 at u = 9 is 18. The gradient of y = (x3 + 1)2 at x = 2 is 216. The chain
rule is used to confirm this.
du
dx= 3x2 and at x = 2
du
dx= 12
dy
du= 2u and at u = 9
dy
du= 18
dy
dx= dy
du· du
dx= 216
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Chapter 22 — Differentiation Techniques 587
Example 7
Find the derivative of y = (3x+ 4)20.
Solution
Let u = 3x + 4 then y = u20
Sodu
dx= 3 and
dy
du= 20u19
∴ dy
dx= dy
du· du
dx
= 20u19 · 3
= 60(3x + 4)19
Example 8
Find the gradient of the curve with equation y = 16
3x2 + 1at the point (1, 4).
Solution
Let u = 3x2 + 1 then y = 16u−1
Sodu
dx= 6x and
dy
du= −16u−2
∴ dy
dx= dy
du· du
dx
= −16u−2 · 6x
= −96x
(3x2 + 1)2
∴ at x = 1 the gradient is−96
16= −6.
Example 9
Differentiate y = (4x3 − 5x)−2.
Solution
Let u = 4x3 − 5x
then y = u−2
anddu
dx= 12x2 − 5 and
dy
du= −2u−3
∴ dy
dx= dy
du· du
dx
= −(2u−3) · (12x2 − 5)
= −2(12x2 − 5)
(4x3 − 5x)3
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588 Essential Mathematical Methods 1 & 2 CAS
Example 10
Use the chain rule to provedy
dx= nxn−1 for y = xn and n a negative integer.
(Assume the result for n a positive integer and x > 0.)
Solution
Let n be a negative integer and y = xn. Then y = 1
x−nas −n is a positive integer.
Let u = x−n then y = 1
u= u−1
Thusdu
dx= −nx−n−1 (−n is a positive integer) and
dy
du= −u−2
∴ dy
dx= dy
du· du
dx
= −u−2 · (−nx−n−1)
= nx−n−1(x−n)−2
= nx−n−1(x2n)
= nxn−1
With function notation the chain rule is stated as:
( f ◦ g)′(x) = f ′(g(x))g′(x)
where f ◦ g(x) = f (g(x)).
Example 11
Given that f (x) = (x2 + 1)3, find f ′(x).
Solution
Now f = k ◦ g where k(x) = x3 and g(x) = x2+ 1
It follows that k′(x) = 3x2 and g′(x) = 2x
By the chain rule f ′(x) = k ′(g(x))g ′(x)
We have in this case f ′(x) = 3(g(x))2 2x
which yields f ′(x) = 6x(x2 + 1)2
Exercise 22B
1 Differentiate each of the following with respect to x :Examples 7, 8, 9
a (x − 1)30 b (x5 − x10)20 c (x − x3 − x5)4
d (x2 + 2x + 1)4 e (x2 + 2x)−2, x �= −2, 0 f
(x2 − 2
x
)−3
, x �= 0
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Chapter 22 — Differentiation Techniques 589
2 a Find the derivative of f : R → R, f (x) = (2x3 + 1)4.
b Find the gradient of f : R → R, f (x) = (2x3 + 1)4 at the point (1, 81).
3 a Find the gradient of the curve with equation y = 1
x + 3at the point
(1, 1
4
).
b Find the gradient of the curve with equation y = 1
(x + 3)3, at the point
(1, 1
64
).
4 The diagram is a sketch graph
of a function with f (x) = 1
2x + 3. x = – –3
2
y
x0
0, –13
a Find the gradient of the
curve at the points(0, 1
3
).
b Find the coordinates of the
points on the curve for which
the gradient is −2
9.
5 The diagram is a sketch graph of
the functions y = 1
xand y = −1
x.
y
x
y = – 1x
y = – – 1x
y = – 1x
P(–1, 1)
Q(–1, –1)
2
1
0 1 2 3 4–1–1
–2
–2–3–4
–a Find the gradient of y = 1
xat the point
(2, 1
2
).
b Without further calculation
state the gradient of y = −1
xat
(2, −1
2
).
c Find the equation of the tangent at the point (1, 1) of y = 1
x.
d Find the equation of the tangent at the point (1, –1) of y = −1
x.
e Find the equations of the tangents at points P and Q and find their point of intersection.
f Draw sketch graphs of y = 1
xand y = −1
xon the same set of axes and draw in the four
tangents.
22.3 Differentiating rational powers(x
p/q)
Using the chain rule in the formdy
du= dy
dx· dx
du
with y = u 1 = dy
dx· dx
dy
and thusdy
dx= 1
dx
dy
, fordx
dy�= 0
Let y = x1n , where n ∈ Z \ {0} and x > 0.
Then yn = x anddx
dy= nyn−1.
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590 Essential Mathematical Methods 1 & 2 CAS
From the above resultsdy
dx= 1
nyn−1
= 1
n
(x
1n
)n−1
= 1
nx
1n −1
For y = x1n ,
dy
dx= 1
nx
1n −1; n ∈ Z \ {0} and x > 0.
This result may now be extended to any rational power.
Let y = xpq , where p, q ∈ Z \ {0}.
Write y =(
x1q
)p
. Let u = x1q . Then y = up.
The chain rule yieldsdy
dx= dy
du· du
dx
= pu p−1 · 1
qx
1q −1
= p
(x
1q
)p−1
· 1
qx
1q −1
= p
qx
pq − 1
q x1q −1
= p
qx
pq −1
Thus the result for any non-zero rational power has been established and, in fact, it is true for
any non-zero real power.
For f (x) = xa, f ′(x) = axa−1, for x > 0 and a ∈ R.
These results have been stated for
x > 0, as (−3)12 is not defined
although (−2)13 is.
The graphs of y = x12 , y = x
13 and
y = x14 are shown.
The domain of each has been taken
to be R +.
1.5
1.0
0.5
–0.5
–0.5
0.5 1.0 1.5 2.0 2.5 3.00
y
x
y = x
y = x
y = x14
13
12
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Chapter 22 — Differentiation Techniques 591
The figure to the right is the graph of the function f : R → R, f (x) = x13 .
Note that the values shown here
are −0.08 ≤ x ≤ 0.08.
From this it can be seen that the
tangent at the origin of y = x13
is on the y-axis.
–0.2
0.4
0.2
–0.4
0.02–0.02–0.04–0.06–0.08 0.04 0.06 0.080
y
x
3y = x
1
Use a calculator to investigate
graphs of this type further.
Example 12
Find the derivative of each of the following with respect to x :
a 4x23 b x
15 − 2x−3
Solution
a Let y = 4x23 , then
dy
dx= 4 × 2
3× x
23 −1 = 8
3x
−13
b Let y = x15 − 2x−3, then
dy
dx= 1
5x
15 −1 − 2 × −3x−3−1
= 1
5x
−45 + 6x−4
Exercise 22C
1 Find the derivative of each of the following with respect to x :Example 12
a x13 b x
32 , x > 0 c x
52 − x
32 , x > 0
d 2x12 − 3x
53 , x > 0 e x− 5
6 , x > 0 f x− 12 − 4, x > 0
2 Find the derivative of each of the following with respect to x :
a√
1 + x2 b 3√x + x2 c (1 + x2)−12 d (1 + x)
13
3 a Find the gradient of y = x13 at each of the following points:
i
(1
8,
1
2
)ii
(−1
8, −1
2
)iii (1, 1) iv (−1, −1)
b Comment on your results.
4 Consider the graphs of y = x12 and y = x
13 for x > 0.
a Find
{x : x
12 < x
13
}.
b Find the values for x for which the gradient of y = x12 is greater than the gradient
of y = x13 .
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592 Essential Mathematical Methods 1 & 2 CAS
5 Differentiate each of the following with respect to x :
a(2 − 5
√x)2
b(3√
x + 2)2
c2 + √
x
x2
dx2 + 2√
xe 3
√x
(x2 + 2
)
22.4 The second derivativeFor f : R →R, f (x) = 2x2 + 4x + 1, the derived function has rule f ′(x) = 4x + 4,
i.e. the derivative of 2x2 + 4x + 1 is 4x + 4.
The second derivative of 2x2+ 4x + 1 is 4, i.e. the derivative of 4x + 4.
The function notation for the second derivative is f ′′(x) = 4.
In the Leibniz notation
if y = 2x2 + 4x + 1,dy
dx= 4x + 4 and the second derivative is written
d2 y
dx2, and
d2 y
dx2= 4.
Example 13
For each of the following find f ′′(x).
a f (x) = 3x3 + 2x−1 + 1, x �= 0 b f (x) = x12 + 3x−4 + 1, x > 0
c f (x) = x4 + x− 32 + 1, x > 0
Solution
a f ′(x) = 9x2 − 2x−2, x �= 0
f ′′(x) = 18x+ 4x−3, x �= 0
b f ′(x) = 1
2x− 1
2 − 12x−5, x > 0
f ′′(x) = −1
4x− 3
2 + 60x−6, x > 0
c f ′(x) = 4x3 − 3
2x− 5
2 , x > 0
f ′′(x) = 12x2 + 15
4x− 7
2 , x > 0
If S(t) denotes the displacement of a body at time t,
then S′(t) gives the velocity at time t,
and S′′(t) gives acceleration at time t.
Exercise 22D
1 Find f ′′(x) for each of the following:Example 13
a f (x) = x3 + 2x + 1 b f (x) = 3x + 2 c f (x) = (3x + 1)4
d f (x) = x12 + 3x3, x > 0 e f (x) = (x6 + 1)3 f f (x) = 5x2 + 6x−1 + 3x
32
2 For each of the following findd2 y
dx2:Example 13
a y = 3x3 + 4x + 1 b y = 6 c y = 6x2 + 3x + 1
d y = (6x + 1)4 e y = (5x + 2)4 f y = x3 + 2x2 + 3x−1
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Chapter 22 — Differentiation Techniques 593
3 The height of a stone is (20t – 4.9t2) metres at a time t seconds after it is thrown. What is
the acceleration?
4 After t seconds, the x-coordinate of a particle moving along the x-axis is given by
x(t) = 4t – 3t3, where the units on the x-axis are in metres.
a Find:
i the x-coordinate after 2 seconds of motion
ii the velocity of the particle at the start
iii the velocity of the particle after half a second
iv the velocity of the particle after 2 seconds.
b When is the acceleration zero?
c What is the average velocity during the first 2 seconds?
22.5 Sketch graphsIn Chapter 20 it was demonstrated how calculus may help us sketch the graphs of polynomial
functions. In this section we show how the same techniques may be applied to non-polynomial
functions.
For example: Let f : R \ {0} → R, f (x) = 1
x2+ x .
This function is not defined at x = 0.
1 Consider the behaviour of the function as the magnitude of x becomes large.
x f (x)
1 2.000000000000000
2 2.250000000000000
4 4.062500000000000
8 8.015625000000000
16 16.003906250000000
32 32.000976562500000
64 64.000244140625000
128 128.000061035156250
256 256.000015258789063
512 512.000003814697266
1024 1024.00000095367432
2048 2048.00000023841858
This section of a spreadsheet demonstrates the
behaviour of increasing x . This has been done
by using increasing powers of 2.
We see that as x → ∞, f (x) → x .
This shows us that the line y = x is an
oblique asymptote for the graph.
It is clear also that as x → −∞, f (x) → x .
The behaviour of the function as the magnitude of x approaches 0 is demonstrated below.
A spreadsheet has been used and the values of x considered are 20, 2−1, 2−2, 2−3, etc.
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594 Essential Mathematical Methods 1 & 2 CAS
a
x f (x)
1.0000000000 2.000000000000
0.5000000000 4.500000000000
0.2500000000 16.250000000000
0.1250000000 64.125000000000
0.0625000000 256.062500000000
0.0312500000 1024.031250000000
0.0156250000 4096.015625000000
0.0078125000 16384.007812500000
0.0039062500 65536.003906250000
Spreadsheet a shows the behaviour as x
approaches 0 from the right.
f (x) becomes increasingly large.
We write limx→0+
f (x) = ∞.
b
x f (x)
–1.0000000000 0.000000000000
–0.5000000000 3.500000000000
–0.2500000000 15.750000000000
–0.1250000000 63.875000000000
–0.0625000000 255.937500000000
–0.0312500000 1023.968750000000
–0.0156250000 4095.984375000000
–0.0078125000 16383.992187500000
–0.0039062500 65535.996093750000
Spreadsheet b shows the behaviour as x
approaches 0 from the left.
Again f (x) becomes increasingly large.
We write limx→0−
f (x) = ∞.
2 Consider the axis intercepts.
The function cannot have a y-axis intercept, as it is not defined at x = 0. It would cross the
x-axis when1
x2+ x = 0, i.e. when x3 = −1, which implies x = −1.
3 Consider points of zero gradient (turning points).
For f (x) = 1
x2+ x we have f ′(x) = −2
x3+ 1 and f ′(x) = 0 implies x = 2
13 , i.e. there is a
point on the curve at x = 213 at which the gradient is zero.
As the graph is not continuous for R, it is advisable to use a gradient chart for x > 0 only.
f ′(x)
213
shape of f
– +0
x
6
4
2
0 2 4 6–2
–2
–4
–4–6
(1.26, 1.89)
y
x
y = –– + x1x2
Thus there is a minimum at the point (1.26, 1.89)
(coordinates correct to 2 decimal places).
The spreadsheet above shows how rapidly the
graph of y = 1
x2+ x moves towards y = x.
At x = 4, f (x) = 4.0625 and at x = 8,
f (x) = 8.015625.
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Chapter 22 — Differentiation Techniques 595
Example 14
For f : R \ {−1} → R, f (x) = x2 + 3
x + 1, sketch the graph.
Solution
By division, f (x) = x − 1 + 4
x + 1.
Behaviour as magnitude of x increases
As x → ∞, f (x) → x – 1, from above.
As x → –∞, f (x) → x – 1, from below.
There is an oblique asymptote with equation y = x − 1.
Behaviour as x→−1
limx→−1+
f (x) = ∞ and limx→−1−
f (x) = −∞
There is a vertical asymptote x = –1.
Axis intercepts
When x = 0, f (x) = 3. There is no x-axis intercept as x2 + 3 �= 0 for all x ∈ R.
Turning points
For f (x) = x − 1 + 4
x + 1
f ′(x) = 1 − 4
(x + 1)2
Where f ′(x) = 0
1 − 4
(x + 1)2= 0
which implies (x + 1)2 = 4
i.e. x + 1 = 2 or x + 1 = –2
∴ x = 1 or x = –3
f (1) = 2 and f (–3) = –6
We consider two gradient charts.
First for x > –1.
1
f ′(x)
shape of f
– +0
x
∴ a minimum at (1, 2)
Next, for x < −1.
+
–3
f ′(x)
shape of f
–0
x
∴ a maximum at (–3, –6)
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596 Essential Mathematical Methods 1 & 2 CAS
We are now in a position to sketch the graph.
y
x
x = –1
y = x – 1
(0, 3)
(–3, –6)
10
0 5 10 15
–5
–10
–15
–15 –10 –5
5
y = ––––––x2 + 3x + 1
Exercise 22E
1 The equation of a curve is y = 4x + 1
x.
a Find the coordinates of the turning points.
b Find the equation of the tangent to the curve at the point where x = 2.
2 Find the x-coordinates of the points on the curve y = x2 − 1
xat which the gradient of the
curve is 5.
3 Find the gradient of the curve y = 2x − 4
x2at the point where the curve crosses the x-axis.
4 For the curve y = x − 5 + 4
x, find:
a the coordinates of the points of intersection with the axes
b the equations of all asymptotes c the coordinates of all turning points.
Use this information to sketch the curve.
5 If x is positive, find the least value of x + 4
x2.
6 For positive values of x, sketch the graph of y = x + 4
xand find the least value of y.
7 Sketch the graphs of each of the following, indicating the coordinates of the axes interceptsExample 14
and the turning point, and the equations of asymptotes.
a y = x + 1
x, x �= 0 b y = 1
x2− x, x �= 0
c y = x + 1 + 1
x + 3, x �= 3 d y = x3 + 243
x, x �= 0
e y = x − 5 + 1
x, x �= 0 f y = x2 − 4
x + 2, x �= −2
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Review
Chapter 22 — Differentiation Techniques 597
Chapter summary
The general result for differentiating functions, including powers of x with negative
integers:
For f (x) = xn, f ′(x) = nxn−1, n a non-zero integer. For f (x) = 1, f ′(x) = 0.
We note that for n ≤ −1, we take the domain of f to be R \ {0}, and for n ≥ 1 we take the
domain of f to be R.
The chain rule is often used to differentiate some more complicated functions by
transforming the original function into two simpler functions:
e.g. f (x) is transformed to h(x) and g(u), which are ‘chained’ together as
xh−→u
g−→y
Using Leibniz notation the chain rule is stated asdy
dx= dy
du· du
dx.
With function notation the chain rule is stated as
( f ◦ g)′(x) = f ′(g(x))g′(x) where ( f ◦ g)(x) = f (g(x))
The general result for any non-zero real power:
For f (x) = xa, f ′(x) = axa − 1, for x > 0 and a ∈ R.
The function for the second derivative is f ′′.
In Leibniz notation it is written asd 2 y
dx2.
When using calculus as an aid to sketching graphs of polynomial and non-polynomial
functions, the following should be considered:
1 The behaviour of the function as the magnitude of x becomes large
2 The axis intercepts
3 The points of zero gradient (turning points).
Multiple-choice questions
1 If f (x) = 4x4 − 12x2
3xthen f ′(x) equals
A16x3 − 24x
3B 4x2 − 4 C
16x3 − 24x
3x
D 4x2 − 8x E8x3 − 16x
3x
2 If f (x) = 2xpq , where p and q are integers, f ′(x) equals
A 2x(p−q)
q B 2pxpq −1
C 2 D2p
qx
(p−q)q E
2p
qx
3 For f : R \ {2} → R, where f (x) = 4 + 4
2 − x, f ′(x) > 0 for
A R \ {2} B R C x < 2 D x > 2 E x > 4
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Rev
iew
598 Essential Mathematical Methods 1 & 2 CAS
4 A particle moves in a straight line so that its position x cm from a fixed point O at time
t s (t ≥ 0) is given by x = −t3 + 7t2 − 14t + 6. The particle’s acceleration at t = 3 is
A −4 cm/s2 B 3 cm/s2 C 4 cm/s2 D 8 cm/s2 E 0 cm/s2
5 Let y = f (g(x)), where g(x) = x3, thendy
dxequals
A 3x2 f ′(x3) B 3x2 f (x3) C 2x f (x) f ′(x3)
D 2 f (x) f ′(x3) E 3x2
6 The graph defined by the rule f (x) = x + 1
xhas a local minimum at (a, f (a)).
The value of a is
A −1 B 2 C −5
2D
5
2E 1
7 Which of the following is not true for the curve of y = f (x), where f (x) = x15 ?
A The gradient is defined for all real numbers.
B The curve passes through the origin.
C The curve passes through the points with coordinates (1, 1) and (−1,−1).
D For x > 0 the gradient is positive.
E For x > 0 the gradient is decreasing.
8 Which of the following is not true for y = f (x), where f (x) = x34 ?
A The maximal domain of the function is R+ ∪ {0}.
B f (x) > x for all x > 1.
C The curve of y = f (x) passes through the points with coordinates (1, 1).
D For x > 0 the gradient of the curve is positive.
E For x > 0 the gradient of the curve is decreasing.
9 The derivative of (5x2 + 2x)n is
A n(10x + 2)(5x2 + 2x)n−1 B (5x2 + 2x)n−1 C (10x + 2)n
D n(5x2 + 2x)n−1 E 10x2n−1 + 2xn−1
10 The graph of the function with rule y = k
2(x2 + 1)has gradient 1 when x = 1.
The value of k is
A 1 B −1 C 4 D −4 E −1
4
Short-answer questions (technology-free)
1 Find the derivative of each of the following with respect to x :
a x−4 b 2x−3 c − 1
3x2d − 1
x4
e3
x5f
x2 + x3
x4g
3x2 + 2x
x2h 5x2 − 2
x
2 Find the derivative of each of the following with respect to x :
a x12 b 3√x c − 2
x13
d x43 e x− 1
3 f x− 13 + 2x
35
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Review
Chapter 22 — Differentiation Techniques 599
3 Differentiate each of the following with respect to x :
a (2x + 3)2 b 2(3x + 4)4 c (3 − 2x)−12 d
1
3 + 2x
e1
(2x − 1)23
f3√
2 + x2g
(2x2 − 3
x2
) 13
4 Find the gradient of each of the following curves at the given point:
a y = √x ; (9, 3) b y = 1
2x + 1; (0, 1) c y = 2
x2;
(4,
1
8
)
d y = 3 + 2
x; (1, 5) e y = √
x + 1; (8, 3) f y = (x2 − 7x − 8)3; (8, 0)
5 Find the coordinates of the point(s) on the curve with equation y = 1
xfor which the
gradient is −4.
6 Find the coordinates of the point(s) on the curve with equation y = √x for which the
gradient is 2.
Extended-response questions
1 A solid circular cylinder has radius r cm and height h cm. It has a fixed volume of 400 cm3.
a Find h in terms of r.
b Show that the total surface area, A cm2, of the cylinder is given by A = 2�r2 + 800
r.
c FinddA
dr.
d Solve the equationdA
dr= 0 for r.
e Find correct to 3 significant figures the minimum surface area of the cylinder.
f Sketch the graph of A against r.
2 A rectangle has sides of length x cm and y cm and the area of the rectangle is 16 cm2.
a Find y in terms of x .
b Show that the perimeter, P cm, is given by P = 2x + 32
x.
c Find the value of x for which the value of P is a minimum and find this value of P.
d Sketch the graph of P against x for x > 0.
3 The area of rectangle OABC is 120 cm2. Let the length of OC be x cm, CZ = 5 cm and
AX = 7 cm.
A
O C
X Y
Z
Ba Find the length of OA in terms of x .
b Find the length of OX in terms of x .
c Find the length of OZ in terms of x .
d Find the area, A cm2, of rectangle OXYZ in terms of x .
e Find the value of x for which the area, A cm2, is a minimum.
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Rev
iew
600 Essential Mathematical Methods 1 & 2 CAS
4 The curve with equation y = √x + 2 meets the x-axis at A and the y-axis at B.
a Find the coordinates of A and B.
b By using the chain rule finddy
dx.
c i Find the gradient of the curve where x = −1.
ii Find the equation of the tangent at the point where x = −1.
iii If the tangent meets the x-axis at C and the y-axis at D, find the distance CD.
d Find the values of x for whichdy
dx< 1.
5 An open rectangular box of height h cm has a horizontal rectangular base with side lengths
x cm and 2x cm. The volume of the box is 36 cm3:
a Express h in terms of x .
b Show that the total surface area of the box is given by A = 2x2 + 108
x.
c Calculate the values of x and h which make the total surface area a minimum.
d Sketch the graph of A against x for x > 0.
6 The prism shown in the diagram has a triangular cross-section. The ‘ends’ of the prism
shown are congruent right-angled triangles with the right angles at C and Z.
A
B
C
X
Y
Z
AX = CZ = BY = y cm, AC = XZ = 3x cm and
CB = ZY = 4x cm
The volume of the prism is 1500 cm3.
a Express y in terms of x .
b Show that the total surface area, S cm2, is
given by S = 12x2 + 3000
x.
c FinddS
dx.
d Find the minimum value of S.