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9 Differentiation 2 – Further
Techniques
Leonhard Euler is considered
to be one of the most
important mathematicians of
all time. He was born on 15
April 1707 in Basel,
Switzerland, and died on 18
September 1783 in St
Petersburg, Russia, although
he spent much of his life in
Berlin. Euler’s mathematical
discoveries are in many
branches of mathematics
including number theory,
geometry, trigonometry,
mechanics, calculus and
analysis. Some of the
best-known notation was
created by Euler including the
notation f(x) for a function, e
for the base of natural logs, i
for the square root of ⫺1, p
for pi, © for summation and
many others. Euler enjoyed his
work immensely, writing in
1741, “The King calls me his
professor, and I think I am the
happiest man in the world.”
Even on his dying day he
continued to enjoy
mathematics, giving a mathematics lesson to his grandchildren and doing some work on
the motion of ballons.
In Chapter 8, the basic concepts of differentiation were covered. However, the only functions
that we differentiated all reduced to functions of the form y ⫽ ax n ⫹ p ⫹ k. In this
chapter, we will meet and use further techniques to differentiate other functions. These
include trigonometric, exponential and logarithmic functions, functions that are given
implicitly, and functions that are the product or quotient of two (or more) functions.

217

9 Differentiation 2 – Further Techniques

9.1 Differentiating trigonometric functions
What is the derivative of sin x?

Using our knowledge of sketching the derived function, we know that the graph must
be of this form:

We can use a calculator to draw the derivative graph as above.
This graph looks very much like the cosine function. We now need to see if it is.
In order to prove this, there are two results that need to be investigated. First, we need
sin h
to consider what happens to
for small values of h. The calculator can be used to
h
investigate this:

It is clear that, as h S 0,

sin h
S 1.
h

Second, we also need to investigate

218

cos h ⫺ 1
for small values of h.
h

9 Differentiation 2 – Further Techniques

It is clear that, as h S 0,

cos h ⫺ 1
S 0.
h

Now we can use differentiation by first principles to find the derivative of f1x2 ⫽ sin x.
f1x ⫹ h2 ⫽ sin1x ⫹ h2
f1x ⫹ h2 ⫺ f1x2 ⫽ sin1x ⫹ h2 ⫺ sin x
⫽ sin x cos h ⫹ cos x sin h ⫺ sin x
⫽ sin x1cos h ⫺ 12 ⫹ cos x sin h
f1x ⫹ h2 ⫺ f1x2
sin x1cos h ⫺ 12
cos x sin h

h
h
h
So lim

hS0

f1x ⫹ h2 ⫺ f1x2
⫽ 0 ⫹ cos x using the above results.
h
f1x ⫹ h2 ⫺ f1x2
⫽ cos x
hS0
h

Hence lim

Therefore if f1x2 ⫽ sin x, f¿1x2 ⫽ cos x.
What about the derivative of cos x?
Examining the graph of the derived function using the calculator, this would appear
to be ⫺sin x.

When dealing with trigonometric
functions it is vital that radians
are used. This is because of the
results that we investigated
above.
In degrees, lim

hS0

sin h
⫽⫽ 11 as
h

seen below.

Again, we can use using differentiation by first principles to find the derivative of
f1x2 ⫽ cos x.
f1x ⫹ h2 ⫽ cos1x ⫹ h2
f1x ⫹ h2 ⫺ f1x2 ⫽ cos1x ⫹ h2 ⫺ cos x

So in degrees, the derivative of
sin x is not cos x.
Therefore, for calculus, we must

⫽ cos x cos h ⫺ sin x sin h ⫺ cos x
⫽ cos x1cos h ⫺ 12 ⫺ sin x sin h
f1x ⫹ h2 ⫺ f1x2
cos x1cos h ⫺ 12
sin x sin h

h
h
h
f1x ⫹ h2 ⫺ f1x2
⫽ 0 ⫺ sin x using the previous results.
hS0
h

So lim

Hence lim

hS0

f1x ⫹ h2 ⫺ f1x2
⫽ ⫺sin x
h

Therefore if f1x2 ⫽ cos x, f¿1x2 ⫽ ⫺sin x.

219

9 Differentiation 2 – Further Techniques

Summarizing:
y ⫽ sin x
dy
⫽ cos x
dx

y ⫽ cos x
dy
⫽ ⫺sin x
dx

It is clear that these are connected (as the two functions themselves are). Starting with
sin x, repeated differentiation gives cos x, ⫺sin x, ⫺cos x and then back to sin x. This
cycle can be remembered by

Differentiate

S
C
⫺S
⫺C

We now have the derivatives of two of the six trigonometric functions. The other four
sin x
functions are all defined in terms of sin x and cos x (remembering tan x ⫽
) and so
cos x
this information provides the derivatives of the other four functions.
Proofs of these require the use of rules that have not yet been covered, and hence these
are to be found later in the chapter. However, the results are shown below.
y ⫽ tan x
dy
⫽ sec2 x
dx

y ⫽ csc x

y ⫽ sec x

dy
⫽ ⫺csc x cot x
dx

dy
⫽ sec x tan x
dx

Example
Find the derivative of y ⫽ cos x ⫺ sec x.
dy
⫽ ⫺sin x ⫺ sec x tan x
dx

Example
Find the derivative of y ⫽ 8 sin x.
dy
⫽ 8 cos x
dx

Exercise 1
Find the derivative of each of these.
1 y ⫽ tan x ⫹ 3

2 y ⫽ sin x ⫺ csc x

3 y ⫽ sin x ⫹ 6x
5 y ⫽ 7 cot x

4 y ⫽ 5 cos x
6 y ⫽ ⫺3 sec x

7 y ⫽ 9x2 ⫺ 4 cos x

8 y ⫽ 7x ⫺ 5 sin x ⫺ sec x

2

220

y ⫽ cot x
dy
⫽ ⫺csc2 x
dx

9 Differentiation 2 – Further Techniques

9.2 Differentiating functions of functions
(chain rule)
The chain rule is a very useful and important rule for differentiation. This allows us to
differentiate composite functions. First consider y ⫽ 1ax ⫹ b2 n.

Investigation
Consider these functions:
1 y ⫽ 12x ⫹ 12 2

2 y ⫽ 12x ⫹ 12 3

3 y ⫽ 13x ⫺ 22 4

4 y ⫽ 13x ⫺ 22 5

5 y ⫽ 14 ⫺ x2 2

6 y ⫽ 14 ⫺ x2 3

Using knowledge of the binomial theorem and differentiation, find the derivatives of the
You should have noticed a pattern that will allow us to take a “shortcut”, which we
always use, when differentiating this type of function.
This is that for functions of the form y ⫽ 1ax ⫹ b2 n,
dy
⫽ an1ax ⫹ b2 n⫺1
dx
This is a specific case of a more general rule, known as the chain rule, which can be
stated as:
dy du
dy
#

dx
du dx

We can consider this as
differentiating the bracket to
the power n and then
multiplying by the derivative
of the bracket.

This is not due to cancelling!

This is where y is a function of a function. This means that we can consider y as a function
of u and u as a function of x.

Proof
Consider y ⫽ g1u2 where u ⫽ f1x2.
If dx is a small increase in x, then we can consider du and dy as the corresponding
increases in u and y.
Then, as dx S 0, du and dy also tend to zero.
We know from Chapter 8 that

dy
dy
⫽ lim ¢ ≤
dx dxS0 dx

⫽ lim ¢

dy du
# ≤
du dx

⫽ lim ¢

dy
du
≤ # lim ¢ ≤
du dxS0 dx

dxS0

dxS0

⫽ lim ¢
duS0

So

dy
du
≤ # lim ¢ ≤
du dxS0 dx

dy
dy du
#

dx
du dx

The use of this rule is made clear in the following examples.

221

9 Differentiation 2 – Further Techniques

Example
Differentiate y ⫽ 13x ⫺ 42 4.
Let u ⫽ 3x ⫺ 4 and y ⫽ u4.
Then

dy
du
⫽ 4u3 and
⫽ 3.
du
dx

Hence

dy
⫽ 4u3 # 3 ⫽ 12u3.
dx

Substituting back for x gives

dy
⫽ 1213x ⫺ 42 3.
dx

Example
Differentiate y ⫽ 15 ⫺ 2x2 7.
Let u ⫽ 5 ⫺ 2x and y ⫽ u7.
dy
du
⫽ 7u6 and
⫽ ⫺2.
du
dx
dy
⫽ 7u6 # ⫺2 ⫽ ⫺14u6.
Hence
dx
dy
⫽ ⫺1415 ⫺ 2x2 6.
Substituting back for x gives
dx
Then

We will now apply the chain rule to other cases of a function of a function.

Example
Differentiate y ⫽ sin 4x.
Let u ⫽ 4x and y ⫽ sin u.
dy
du
⫽ cos u and
⫽ 4.
du
dx
dy
Hence
⫽ 4 cos u.
dx
dy
⫽ 4 cos 4x.
Substituting back for x gives
dx
Then

Example
Differentiate y ⫽ cos2 13x2.
Remember that this means y ⫽ 1cos 3x2 2.
Here there is more than one composition and so the chain rule must be extended
to:
dy
dy du dv
# #

dx
du dv dx
Let v ⫽ 3x and u ⫽ cos v and y ⫽ u2.

222

9 Differentiation 2 – Further Techniques

dy
dv
du
⫽ 2u,
⫽ ⫺sin v and
⫽ 3.
du
dv
dx
dy
⫽ ⫺6u sin v.
Hence
dx
dy
⫽ ⫺6 cos 3x sin 3x.
Substituting back for x gives
dx
Then

This is the formal version of the working for chain rule problems. In practice, the
substitution is often implied, as shown in the following examples. However, it is
important to be able to use the formal substitution, both for more difficult chain rule
examples and as a skill for further techniques in calculus.

Example
f1x2 ⫽ 217 ⫺ 3x2 6 ⫺ tan 2x
f¿1x2 ⫽ 1217 ⫺ 3x2 5 # 1⫺32 ⫺ 2 sec2 2x

This working is sufficient and
is what is usually done.

⫽ ⫺3617 ⫺ 3x2 5 ⫺ 2 sec2 2x

Example
f1x2 ⫽ tan13x2 ⫺ 42 ⫹

2
22x ⫺ 1

⫽ tan13x2 ⫺ 42 ⫹ 212x ⫺ 12 ⫺ 2
1

1
3
f¿1x2 ⫽ 3sec2 13x2 ⫺ 42 # 6x4 ⫹ B2 # ⫺ 12x ⫺ 12 ⫺ 2 # 2R
2
2
⫽ 6x sec2 13x2 ⫺ 42 ⫺
3
12x ⫺ 12 2

Exercise 2
Differentiate the following:
1 f1x2 ⫽ 1x ⫹ 42 2

2 f1x2 ⫽ 12x ⫹ 32 2

3 f1x2 ⫽ 13x ⫺ 42 2

4 f1x2 ⫽ 15x ⫺ 22 4

5 f1x2 ⫽ 15 ⫺ x2 3

6 f1x2 ⫽ 17 ⫺ 2x2 4

7 y ⫽ 19 ⫺ 4x2 5

8 y ⫽ 412x ⫹ 32 6

9 y ⫽ 13x ⫹ 82 2

5

3
11 y ⫽ 26x
⫺5

10 y ⫽ 12x ⫺ 92 3
13 f1x2 ⫽
16 N ⫽

4
5x ⫺ 4
5

218 ⫺ 5p2 3

1
19 y ⫽ ⫺sin x
2

14 f1x2 ⫽

7
3 ⫺ 8x

1

12 y ⫽

1

23x ⫺ 2
3
15 P ⫽
14 ⫺ 3k2 2

17 y ⫽ sin 4x

18 y ⫽ cos 3x

20 y ⫽ tan 6x

21 y ⫽ sec 9x

223

9 Differentiation 2 – Further Techniques

22 y ⫽ 6x ⫹ cot 3x
24 y ⫽ sin 5x ⫺

23 y ⫽ csc 2x ⫹ 13x ⫹ 22 4
4

213x ⫹ 42

27 y ⫽ 3x4 ⫺ cos3 x
29 y ⫽ cos ¢3x ⫺

p

4

5

25 y ⫽ sin3 x
28 y ⫽

26 y ⫽ tan2 14x2

2
⫺ sec2 12x2
13x ⫺ 42 5

30 y ⫽ tan1 2x ⫹ 12

9.3 Differentiating exponential and
logarithmic functions
Investigation
Draw graphs of: (a) y ⫽ 10x (b) y ⫽ 5x (c) y ⫽ 3x (d) y ⫽ 2x
For each graph, draw the derivative graph, using a graphing calculator.
You should notice that the derivative graph is of a similar form to the original, that is, it
is an exponential graph.
For y ⫽ 3x, the derivative graph is just above the original.

For y ⫽ 2x, the derivative graph is below the original.

This suggests that there is a function for which the derivative graph is identical to the
original graph and that the base of this function lies between 2 and 3. What is this base?
This question was studied for many years by many mathematicians including Leonhard
Euler, who first used the symbol e. The answer is that this base is e. Check that y ⫽ ex
produces its own graph for the derived function on your calculator. Remember that
e ⫽ 2.71828 p is an irrational number.

224

9 Differentiation 2 – Further Techniques

During the study of exponential functions in Chapter 5, we met the natural exponential
function, y ⫽ ex. The significance of this function becomes clearer now: the derivative
of ex is itself.
d x
1e 2 ⫽ ex
dx

Below is a formal proof of this.

Proof of derivative of ex
Consider the curve y ⫽ ax.
y

y ⫽ ax

B

A

1

␦x

Gradient of the chord AB ⫽
Gradient at A ⫽ lim ¢
dxS0

x

dx

dx

Now consider two general points on the exponential curve.
y ⫽ ax

y

D

C

1
x

x ⫹ ␦x

x

␦x

ax⫹dx ⫺ ax
x ⫹ dx ⫺ x
dxS0
dx

Hence the gradient at C is ax multiplied by the gradient at A.
But when a ⬅ e, the gradient at A ⫽ 1 (this can be checked on a graphing calculator).
Gradient at C ⫽ ex # 1
d x
Hence
1e 2 ⫽ ex
dx

This is another property of the
curve y ⫽ ex . At (0,1) its

225

9 Differentiation 2 – Further Techniques

Example
Differentiate y ⫽ e5x.
This can be considered as y ⫽ eu where u ⫽ 5x.
Since

dy
du
⫽ 5 and
⫽ eu
dx
du

dy
⫽ eu # 5
dx
⫽ 5e5x

Example
y ⫽ ex ⫺1
2

1

dy
2
⫽ 2xex ⫺1
dx

The result for ex can be combined with the chain rule to create a general rule for
differentiating exponential functions.
d f 1x2
1e 2 ⫽ f¿1x2ef 1x2
dx
This now allows us to differentiate the inverse function of y ⫽ ex, known as the
natural logarithmic function y ⫽ ln x.
As y ⫽ ex and y ⫽ ln x are inverse functions, from Chapter 5 we know that eln x ⫽ x.
We can differentiate both sides of this equation (with respect to x).
d ln x
d
1e 2 ⫽
1x2
dx
dx
1 eln x #

d
1ln x2 ⫽ 1
dx

1

d
1
1ln x2 ⫽ ln x
dx
e

1

d
1
1ln x2 ⫽
x
dx

This is a very important result.
d
1
1ln x2 ⫽
x
dx
The result for ln x can be combined with the chain rule to create this general result:
If
Then

226

y ⫽ ln1f1x2 2
dy
1 #

f¿1x2
dx
f1x2
f¿1x2

f1x2

This result is particularly
important for integration in
Chapter 15.

9 Differentiation 2 – Further Techniques

Example
Differentiate y ⫽ ln14x2.
dy
1
1
⫽4#

x
dx
4x

Example
Differentiate y ⫽ ln1sin x2.
dy
1 #

cos x
dx
sin x
⫽ cot x

Example
Differentiate y ⫽ ln13x ⫺ 22.
dy
1
#3

dx
3x ⫺ 2
3

3x ⫺ 2

Using the results for ex and ln x helps us generalize so that we can find the derivatives
of any exponential or logarithmic function.
To find out how to differentiate y ⫽ ax we first consider y ⫽ 4x.
Since eln 4 ⫽ 4, we can rewrite this function as y ⫽ 1eln 4 2 x.
y ⫽ ex ln 4
1

dy
⫽ ln 4ex ln 4
dx

1

dy
⫽ ln 4 # 4x
dx

In general, ax ⫽ 1eln a 2 x ⫽ ex ln a
and so

dy
⫽ ln a # ex ln a ⫽ ln a # ax
dx

d x
1a 2 ⫽ ln a # ax
dx

We will now look at y ⫽ loga x. In this case the change of base formula will help.

227

9 Differentiation 2 – Further Techniques

loga x ⫽

ln x
1 #

ln x
ln a
ln a

Differentiating gives us
d
1 #1
1loga x2 ⫽
dx
ln a x

1
x ln a
1
d
1loga x2 ⫽
dx
x ln a

Considering these two general results, it is clear that the results for ex and ln x are
actually just special cases.

Example
Differentiate y ⫽ 3 log7 x.
dy
3

dx
x ln 7

Example
Differentiate y ⫽ k # 62x.
dy
⫽ 2k ln 6 # 62x
dx

Sometimes it is useful to use laws of logarithms to assist the differentiation.

Example
Differentiate y ⫽ ln¢

2x ⫹ 1
≤.
x⫺4

We can consider this as y ⫽ ln12x ⫹ 12 ⫺ ln1x ⫺ 42.
So

228

dy
2
1

dx
2x ⫹ 1
x⫺4
21x ⫺ 42 ⫺ 12x ⫹ 12

12x ⫹ 12 1x ⫺ 42
⫺9

12x ⫹ 12 1x ⫺ 42

9 Differentiation 2 – Further Techniques

Exercise 3
Differentiate the following:
1 f1x2 ⫽ e3x
2
4 f1x2 ⫽ 5x
e
7 f1x2 ⫽ e2x⫹3
10 f1x2 ⫽ ⫺2 ln 4x

2 f1x2 ⫽ e7x

3 f1x2 ⫽ ⫺e4x

6
e9x
8 f1x2 ⫽ ln 3x

6 f1x2 ⫽ ex

5 f1x2 ⫽ ⫺

11. f1x2 ⫽ ln12x2 ⫹ 42

2

9 f1x2 ⫽ ln 7x
12 y ⫽ 4x

13 y ⫽ 10x

14 y ⫽ 6 # 5x

16 y ⫽ ln 2x ⫺ 2x

17 y ⫽ log2 x

18 y ⫽ log8 x

19 y ⫽ 4x ⫺ log5 x

20 y ⫽ e4x ⫺ sin 2x ⫹ ln x

21 y ⫽ ln1cos x2

22 y ⫽ ln1tan x2

23 y ⫽ tan1ln x2

15 y ⫽ e3x ⫺ 3x

9.4 Product rule
Using the chain rule, y ⫽ 12x ⫹ 32 4 can be differentiated without multiplying out the
brackets first. However, this does not really help to differentiate y ⫽ 13x ⫹ 22 12x ⫺ 32 3
without some unpleasant simplification. Equally, we cannot currently differentiate
y ⫽ ex sin x. These functions are products of two functions, and to be able to
differentiate these we need to use the product rule.
For y ⫽ uv where u and v are functions of x,
dy
du
dv
⫽v
⫹u
dx
dx
dx

This is sometimes
remembered in the shortened
dy
form
⫽ v du ⫹ u dv
dx

Proof
Consider y ⫽ uv where u and v are functions of x.
If dx is a small increase in x, and du, dv and dy are the corresponding increases in u,
v and y, then
y ⫹ dy ⫽ 1u ⫹ du2 1v ⫹ dv2 ⫽ uv ⫹ udv ⫹ vdu ⫹ dudv
As y ⫽ uv, dy ⫽ udv ⫹ vdu ⫹ dudv
dy
dv
du
dv
⫽u ⫹v
⫹ du
dx
dx
dx
dx

So

Now when dx S 0,
Therefore

dy
dy du
du dv
dv
S ,
S ,
S , du S 0
dx
dx dx
dx dx
dx

dy
dy
⫽ lim ¢ ≤
dx dxS0 dx

1

dy
dv
du
⫽u ⫹v
⫹0
dx
dx
dx

1

dy
dv
du
⫽u ⫹v
dx
dx
dx

229

9 Differentiation 2 – Further Techniques

Example
Differentiate y ⫽ 13x ⫹ 22 12x ⫺ 32 3.
Let y ⫽ uv where u ⫽ 3x ⫹ 2 and v ⫽ 12x ⫺ 32 3.
du
dv
⫽3
⫽ 312x ⫺ 32 2 # 2
dx
dx
⫽ 612x ⫺ 32 2
dy
⫽ 13x ⫹ 22 # 612x ⫺ 32 2 ⫹ 312x ⫺ 32 3
dx
⫽ 312x ⫺ 32 2 3213x ⫹ 22 ⫹ 2x ⫺ 34
⫽ 312x ⫺ 32 2 18x ⫹ 12

There are often common
factors which can be used to

Example
Differentiate y ⫽ ex sin x.
Let y ⫽ uv where u ⫽ ex and v ⫽ sin x.
du
⫽ ex
dx

This is the mechanics of the
solution. It is not absolutely
necessary for it to be shown as
part of the solution.

dv
⫽ cos x
dx

dy
⫽ ex sin x ⫹ ex cos x
dx
⫽ ex 1sin x ⫹ cos x2

Example
Differentiate y ⫽ 4x2 ln x.
Let y ⫽ uv where u ⫽ 4x2 and v ⫽ ln x.
du
⫽ 8x
dx

1
dv

x
dx

dy
1
⫽ 8x ln x ⫹ 4x2 #
x
dx
⫽ 8x ln x ⫹ 4x
⫽ 4x12 ln x ⫹ 12

Example
Differentiate y ⫽ 12x ⫹ 12 3e2x cos x.
This example is a product of three functions. We need to split it into two parts
and then further split the second part.
Let y ⫽ uv where u ⫽ 12x ⫹ 12 3 and
du
⫽ 312x ⫹ 12 2 # 2
dx
⫽ 612x ⫹ 12 2

230

v ⫽ e2x cos x.
dv
For
we need to use the product
dx
rule again.

9 Differentiation 2 – Further Techniques

Let v ⫽ fg where f ⫽ e2x and g ⫽ cos x
dg
df
⫽ 2e2x
⫽ ⫺sin x
dx
dx
dv
⫽ 2e2x cos x ⫺ e2x sin x
dx
⫽ e2x 12 cos x ⫺ sin x2
dy
⫽ 612x ⫹ 12 2e2x cos x ⫹ 12x ⫹ 12 3e2x 12 cos x ⫺ sin x2
dx
⫽ 12x ⫹ 12 2e2x 36 cos x ⫹ 12x ⫹ 12 12 cos x ⫺ sin x2 4

Exercise 4
Find the derivative of each of these.
1 y ⫽ x2 sin x

2 y ⫽ x3 cos x

3 y ⫽ 3x2ex

4 y ⫽ e3x sin x

5 y ⫽ ln x sin x

6 y ⫽ sin x cos x

7 y ⫽ sin 3x cos 2x

8 y ⫽ x2 1x ⫺ 12 2

9 y ⫽ x3 1x ⫺ 22 4

10 y ⫽ 2x3 13x ⫹ 22 2

11 y ⫽ 1x ⫺ 22 12x ⫹ 12 3

12 y ⫽ 1x ⫹ 52 2 13x ⫺ 22 4

13 y ⫽ 15 ⫺ 2x2 3 13x ⫹ 42 2

14 y ⫽ 13x ⫹ 42 3 sin x

15 y ⫽ 5x cos x

16 y ⫽ x3 log6 x

17 y ⫽ e4x sec 3x

18 y ⫽ 12x ⫺ 12 3 csc 3x

19 y ⫽ 4x log8 x

20 y ⫽ x ln12x ⫹ 32

21 y ⫽ 4x2 ln1x2 ⫹ 2x ⫹ 52

22 y ⫽ e3x sec ¢2x ⫺

p

4

23 y ⫽

3
p
tan ¢3x ⫹ ≤
4
2
x

25 y ⫽ e3x 1x ⫹ 22 2 tan x

24 y ⫽ x2 ln x sin x

9.5 Quotient rule
This rule is used for differentiating a quotient (one function divided by another) such as
u
y⫽ .
v
Consider the function y ⫽

13x ⫺ 42 2
x2

⫽ x ⫺2 13x ⫺ 4 2 2

We can differentiate this using the product rule.
Let y ⫽ uv where u ⫽ 13x ⫺ 42 2 and v ⫽ x⫺2.
du
⫽ 613x ⫺ 42
dx

dv
⫽ ⫺2x⫺3
dx

dy
⫽ 6x⫺2 13x ⫺ 42 ⫺ 2x⫺3 13x ⫺ 42 2
dx

231

9 Differentiation 2 – Further Techniques

⫽ 2x⫺3 13x ⫺ 42 33x ⫺ 13x ⫺ 42 4
⫽ 2x⫺3 13x ⫺ 42 142

813x ⫺ 42
x3

In this case it was reasonably easy to rearrange into a product, but that is not always so.
Generally, it is not wise to use the product rule for differentiating quotients as it often
leads to an answer that is difficult to simplify, and hence a rule for differentiating
quotients would be useful.
Consider y ⫽

u
where u and v are functions of x.
v

This can be written as y ⫽ u #

1.
v

d 1
d
¢ ≤ ⫽ ⫺v⫺2 # 1v2
dx v
dx
⫽⫺

This is a slightly different
application of the chain rule.

1 # dv
v2 dx

Using the product rule:
dy
1 du
1 dv
⫽ #
⫹u#⫺ 2#
v dx
dx
v dx

v # du
u dv
⫹⫺ 2#
2
v dx
v dx

du
dv
⫺u
dx
dx

2
v
v

This is the quotient rule:

du
dv
v
⫺u
dy
dx
dx

dx
v2

Example
Differentiate y ⫽

x2

using the quotient rule.

Let u ⫽ 13x ⫺ 42 2

v ⫽ x2

du
⫽ 613x ⫺ 42
dx

dv
⫽ 2x
dx

So

232

13x ⫺ 42 2

This is often remembered as
dy
v du ⫺ u dv

dx
v2
The numerator of the quotient
rule is very similar to the
product rule but the sign is
different.

6x2 13x ⫺ 42 ⫺ 2x13x ⫺ 42 2
dy

dx
x4
2x13x ⫺ 42 33x ⫺ 13x ⫺ 42 4

x4

v2 ⫽ x4

Once again, this is the
mechanics of the solution and
so does not necessarily need
to be shown.

9 Differentiation 2 – Further Techniques

213x ⫺ 42 142

x3
813x ⫺ 42

x3

Example
Differentiate y ⫽
Let u ⫽ e2x

v ⫽ sin x

du
⫽ 2e2x
dx
So

e2x .
sin x
v2 ⫽ sin2 x

dv
⫽ cos x
dx

dy
2e2x sin x ⫺ e2x cos x

dx
sin2 x

e2x 12 sin x ⫺ cos x2
sin2 x

Example
Differentiate y ⫽

ln x
.
12x ⫺ 52 3

Let u ⫽ ln x

v ⫽ 12x ⫺ 52 3

du
1

x
dx

dv
⫽ 612x ⫺ 52 2
dx

v2 ⫽ 12x ⫺ 52 6

1
12x ⫺ 52 3 ⫺ 6 ln x12x ⫺ 52 2
dy
x
So

dx
12x ⫺ 52 6
1
12x ⫺ 52 ⫺ 6 ln x
x

12x ⫺ 52 4

2x ⫺ 5 ⫺ 6x ln x
x12x ⫺ 52 4

Exercise 5
Use the quotient rule to differentiate these.
1 f1x2 ⫽

ex
cos x

2 f1x2 ⫽

6x2
x⫹3

3 f1x2 ⫽

7x
tan x

4 f1x2 ⫽

ln x
4x

5 f1x2 ⫽

ex
x⫺4

6 f1x2 ⫽

x⫹3
x⫺3

233

9 Differentiation 2 – Further Techniques

7 f1x2 ⫽

2x ⫹ 9
x2

8 f1x2 ⫽

4x
1x

9 f1x2 ⫽

10 y ⫽

e3x
9x2

11 y ⫽

log6 x
x⫹6

12 y ⫽

13 y ⫽

ex
e ⫺ e⫺x

14 y ⫽

sin 2x
e6x

15 y ⫽

x

x2e3x
17 y ⫽
1x ⫹ 52 2

x sin x
16 y ⫽
ex

2x
2x ⫺ 1

ln x
ln1x ⫺ 42
413x ⫺ 22 5
12x ⫹ 32 3
sec ¢x ⫹

18 y ⫽

p

4

e2x

p

3
19 y ⫽
ln13x ⫹ 12
cot ¢2x ⫺

20 Use the quotient rule to prove the results for tan x, csc x, sec x and cot x. You need
sin x
to remember that tan x ⫽
.
cos x

Also consider how you could
have proved these two results
using only the chain rule.

9.6 Implicit differentiation
This is the differentiation of functions that are stated implicitly. Until now we have
mostly considered functions that are stated explicitly, that is, y ⫽ p
Functions defined implicitly have equations that are not in the form y ⫽ p Some of
these equations are easily made explicit (such as 2x ⫹ 3y ⫽ 5 ) but others are more
difficult to rearrange. Some of these implicit equations may be familiar, such as the circle
equation 1x ⫺ 42 2 ⫹ 1y ⫹ 32 2 ⫽ 36. Differentiating implicit functions does not require
any further mathematical techniques than those covered so far. The key concept utilized
in implicit differentiation is the chain rule.

Method for implicit differentiation
1. Differentiate each term, applying the chain rule to functions of the variable.
dy
.
2. Rearrange the answer to the form
dx

Example
Find

dy
d2y
and
for 3x2 ⫹ y2 ⫽ 7
dx
dx2

Differentiating with respect to x:
6x ⫹ 2y
1 2y

234

It is possible to rearrange this
function to an explicit form.
However, unless you are told
otherwise, it is often better to
leave it in this form and
differentiate implicitly.

dy
⫽0
dx

Applying the chain rule gives

dy
⫽ ⫺6x
dx

dy
d 2
1y 2 ⫽ 2y # .
dx
dx

9 Differentiation 2 – Further Techniques

1

dy
dx

⫽⫺

both x and y.

3x
y

We can now find the second derivative by differentiating this again.

In a case like this it is important

Using the quotient rule:
⫺3y ⫹ 3x

2

dy
dx2

to be able to explicitly state
dy
⫽ ... so that the second
dx
derivative can be found, but this

dy
dx

y2
⫺3y ⫹ 3x ¢⫺

y2
⫺3y ⫺

is not always the situation.

3x

y

9x2
y

y2
⫺3y2 ⫺ 9x2
y3

d 2y
We would usually leave the answer in this form. However, if we wanted
as
dx2
a function of x, we could proceed as follows:
d 2y
dx2

⫺317 ⫺ 3x2 2 ⫺ 9x2
3

17 ⫺ 3x2 2 2

We know that
3x2 ⫹ y2 ⫽ 7
1 y2 ⫽ 7 ⫺ 3x2

⫺ 21
3

17 ⫺ 3x2 2 2

Example
Find

dy
for 6 sin x ⫺ e3xy3 ⫽ 9.
dx

Differentiating with respect to x:
6 cos x ⫺ ¢3e3xy3 ⫹ e3x3y2

dy
≤⫽0
dx

Use the product rule to
differentiate e3xy3 .

dy
≤ ⫽ 6 cos x
dx
dy
2 cos x
1y⫹
⫽ 3x 2
dx
e y

1 3e3xy2¢y ⫹

1

dy
2 cos x
⫽ 3x 2 ⫺ y
dx
e y

235

9 Differentiation 2 – Further Techniques

Example
Find

dQ
2Q
for sin pp ⫹
⫽ ln p.
dp
1p ⫹ 32 2

Differentiating with respect to p:
dQ
2
1p ⫹ 32 2 ⫺ 21p ⫹ 32 # 2Q
dp
1
p cos pp ⫹

4
1p ⫹ 32
p
dQ
2
1p ⫹ 32 2 ⫺ 4Q1p ⫹ 32
dp
1
1
⫽ ⫺ p cos pp
1p ⫹ 32 4
p
dQ
2
dp
4Q
1
1
⫽ ⫺ p cos pp
2 ⫺
3
1p ⫹ 32
p
1p ⫹ 32
dQ
dp
1
4Q
⫺ p cos pp ⫹
1
2 ⫽
1p ⫹ 32
p
1p ⫹ 32 3
2

12

1p ⫹ 32 2
dQ
4Q

⫺ p1p ⫹ 32 2 cos pp ⫹
dp
p
p⫹3

1

1p ⫹ 32 2
dQ
p
2Q

⫺ 1p ⫹ 32 2 cos pp ⫹
dp
2p
2
p⫹3

Example
Find the equations of the tangents to 3x2y ⫺ y2 ⫽ 27 when x ⫽ 2.
Differentiating with respect to x:
dy
dy
6xy ⫹ 3x2 ⫺ 2y ⫽ 0
dx
dx
dy 2
1 13x ⫺ 2y2 ⫽ ⫺6xy
dx
dy
⫺6xy
1
⫽ 2
dx
3x ⫺ 2y
To find

dy
we now require the y-coordinates. So, from the formula
dx

3x2y ⫺ y2 ⫽ 27, when x ⫽ 2, we find
12y ⫺ y2 ⫽ 27
1 y2 ⫺ 12y ⫹ 27 ⫽ 0
1 1y ⫺ 32 1y ⫺ 92 ⫽ 0
1 y ⫽ 3, y ⫽ 9
At (2, 3)
dy
⫺6 # 2 # 3

dx
12 ⫺ 6
⫺36

6
⫽ ⫺6

236

At (2 , 9)
dy
⫺6 # 2 # 9

dx
12 ⫺ 18
⫺108

⫺6
⫽ 18

9 Differentiation 2 – Further Techniques

So the equation of the tangent is
y ⫺ 3 ⫽ ⫺61x ⫺ 22
1 y ⫽ ⫺6x ⫹ 15

So the equation of the tangent is
y ⫺ 9 ⫽ 181x ⫺ 22
1 y ⫽ 18x ⫺ 27

Some questions will require the second derivative to be found, and a result to be
shown to be true that involves

d2y dy
,
and y. In the examples so far we have found
dx2 dx

dy
d2y
⫽ ... and then differentiated this again with respect to x to find
. With other
dx
dx2
questions, it is best to leave the result as an implicit function and differentiate for a
second time, implicitly. The following two examples demonstrate this.

Example
Show that x2

d2y
dx2

⫹ 4x

dy
1
⫹ 2y ⫽ ⫺ for x2y ⫹ x ln x ⫽ 6x.
x
dx

Differentiating with respect to x:
dy
1
2xy ⫹ x2 ⫹ ln x ⫹ # x ⫽ 6
x
dx
dy
1 2xy ⫹ x2 ⫹ ln x ⫹ 1 ⫽ 6
dx
Differentiating again with respect to x:
2y ⫹ 2x

dy
d2y
dy
1
⫹ 2x ⫹ x2 2 ⫹ ⫽ 0
x
dx
dx
dx
1 x2

d2y
2

dx

⫹ 4x

dy
1
⫹ 2y ⫽ ⫺
x
dx

Example
Given that exy ⫽ cos x, show that 2y ⫹ 2

dy
d2y
⫹ 2 ⫽ 0.
dx
dx

Differentiating with respect to x:
dy
exy ⫹ ex ⫽ ⫺sin x
dx
Differentiating again with respect to x:
exy ⫹ ex

dy
dy
d2y
⫹ ex ⫹ ex 2 ⫽ ⫺cos x
dx
dx
dx

From the original function, ⫺cos x ⫽ ⫺exy
So we have exy ⫹ ex

dy
dy
d2y
⫹ ex ⫹ ex 2 ⫽ ⫺exy
dx
dx
dx

237

9 Differentiation 2 – Further Techniques

1 exy ⫹ 2ex
1 2exy ⫹ 2ex

dy
d2y
⫹ ex 2 ⫽ ⫺exy
dx
dx
dy
d2y
⫹ ex 2 ⫽ 0
dx
dx

Dividing by ex (since ex ⫽ 0 ∀ x H ⺢ ):
1 2y ⫹ 2

d2y
dy
⫹ 2⫽0
dx
dx

Exercise 6
1 Find

dy
for:
dx

a x3 ⫹ xy ⫽ 4

b 4x2 ⫹ y2 ⫽ 9

c y3 ⫺ 1x ⫽ 0

d 1x ⫹ 32 1y ⫹ 22 ⫽ ln x

e xy ⫽ y2 ⫺ 7

f ey ⫽ 1x ⫺ y2 2

g e y ⫽ 9 ⫺ sin 3x

h y ⫽ cos1x ⫹ y2

i x4 ⫽ y ln y

j 1x ⫹ y2 3 ⫽ ey

k

2x 3

2 Find

1x ⫹ y2 4
⫽ 8x ⫺ ex
y

dy
d2y
and
for:
dx
dx2

a 4y ⫹ 3y2 ⫽ x2

b 4xy ⫽ sin x ⫺ y

c xey ⫽ 8

3 For the function defined implicitly by x4 ⫺ 2xy ⫹ y2 ⫽ 4, find the equations
of the tangents at x ⫽ 1.
dy
d2y
⫹ 2 ⫽ 0 for exy ⫽ sin x.
dx
dx
dy
d2y
5 Given that xy ⫽ sin x, show that x2 2 ⫹ 2x ⫹ x2y ⫽ 0.
dx
dx
4 Show that 2y ⫹

6 Show that x3

d2y
dx2

⫹ x2

dy
⫹ xy ⫽ ⫺2 for xy ⫽ ln x.
dx

7 Given that e2xy ⫽ ln x, show that x2e2x ¢4y ⫹ 4

dy
d2y
⫹ 2 ≤ ⫽ ⫺1.
dx
dx

9.7 Differentiating inverse trigonometric
functions
In order to find the derivative of sin⫺1 x (or arcsin x), we apply implicit differentiation.
Consider y ⫽ sin⫺1 x
1 x ⫽ sin y
Differentiating with respect to x:

238

9 Differentiation 2 – Further Techniques

1 ⫽ cos y #

dy
dx

This is because sin y ⫽ x

1 sin2 y ⫽ x2

1

dy
1

dx
cos y

1 1 ⫺ cos2 y ⫽ x2

1

dy
1

dx
21 ⫺ x2

1 cos2 y ⫽ 1 ⫺ x2
1 cos y ⫽ 21 ⫺ x2
For y ⫽ sin⫺1 x

dy
1

dx
21 ⫺ x2

x
We can now consider y ⫽ sin⫺1¢ ≤
a
1

x
⫽ sin y
a

1 x ⫽ a sin y
Differentiating with respect to x:
1 ⫽ a cos y
1

dy
1

dx
a cos y

1

dy

dx

1
a

C

dy
dx

1⫺

This is because sin y ⫽
x2
a2

1 cos2 y ⫽ 1 ⫺

2a2
a 2a2 ⫺ x2 º
1

x2
a2

2a2 ⫺ x2

x
For y ⫽ sin ⫺1 a b
a

Similarly

x
x2
2
a 1 1 ⫺ cos y ⫽ a2

dy
1

2
dx
2a ⫺ x2

dy
x
can be obtained for y ⫽ cos⫺1 1x2 and y ⫽ cos⫺1¢ ≤.
dx
a
For y ⫽ cos ⫺1 x

dy
⫺1

dx
21 ⫺ x2

x
For y ⫽ cos ⫺1 a b
a

dy
⫺1

dx
2a2 ⫺ x2

Now consider y ⫽ tan⫺1 1x 2
1 x ⫽ tan y

239

9 Differentiation 2 – Further Techniques

Differentiating with respect to x:
1 ⫽ sec2 y #
1

dy
dx

Remember that
sec2 x ⫽ tan2 x ⫹ 1

dy
1

dx
sec2 y

So sec2 y ⫽ tan2 y ⫹ 1
⫽ x2 ⫹ 1

dy
1
1
⫽ 2
dx
x ⫹1

For y ⫽ tan ⫺1x

dy
1

2
dx
1⫹x

x
A similar result can be found for tan⫺1¢ ≤.
a

x
For y ⫽ tan⫺1¢ ≤
a

dy
a
⫽ 2
dx
a ⫹ x2

Example
x
Differentiate y ⫽ cos⫺1¢ ≤.
3
dy
⫺1

dx
29 ⫺ x2

Example
Differentiate y ⫽ sin⫺1 14x2.
dy

dx

240

1
1
⫺ x2
B 16
1
1
21 ⫺ 16x2
B 16
4
21 ⫺ 16x2

In this case a ⫽

1.
4

9 Differentiation 2 – Further Techniques

We could also consider these examples to be applications of the chain rule. This may be
easier and shorter (but both methods are perfectly valid). This is demonstrated below.
y ⫽ sin⫺1 14x2
Then

dy
1
#4

dx
21 ⫺ 14x2 2

4
21 ⫺ 16x2

In some cases it is not possible to use the stated results, and the chain rule must be
applied.

Example
Differentiate y ⫽ tan⫺1 1x.
Here we must use the chain rule.
dy
1
# 1 x⫺ 12

2
dx
1 ⫹ 1 1x2 2

1
21x11 ⫹ x2

Exercise 7
Differentiate the following functions.
x
1 y ⫽ sin⫺1 ¢ ≤
5
x
2 y ⫽ cos⫺1¢ ≤
8
3 y ⫽ tan⫺1¢
4 y ⫽ sin⫺1 ¢

x

10

2x

3

5 y ⫽ cos⫺1 13x2
ex
6 y ⫽ tan⫺1¢ ≤
2
7 y ⫽ cos⫺1 2x ⫹ 4
8 y ⫽ tan⫺1 12x ⫺ 12
9 y ⫽ sin⫺1 1ln 5x2

241

9 Differentiation 2 – Further Techniques

9.8 Summary of standard results
This chapter has covered a variety of techniques including the chain rule, product rule,
quotient rule and implicit differentiation. These have produced a number of standard
results, which are summarized below.

y⫽

dy
dx

sin x

cos x

cos x

⫺sin x

tan x

sec2 x

csc x

⫺csc x cot x

sec x

sec x tan x

cot x

⫺csc2 x

ex

ex

ln x

1
x

ax

ax ln a

loga x

1
x ln a

sin⫺1 1x2

cos⫺1 1x2

tan⫺1 1x2
x
sin⫺1¢ ≤
a

1
21 ⫺ x2
⫺1
21 ⫺ x2
1
1 ⫹ x2
1
2a ⫺ x2
2

⫺1

x
cos⫺1¢ ≤
a

2a2 ⫺ x2

x
tan⫺1¢ ≤
a

a
a ⫹ x2
2

Within this chapter and Chapter 8, we have covered all of the differentiation techniques
and skills for IB Higher Level. One of the key skills in an examination is to be able to
identify which technique is required to solve a particular problem. Exercise 8 contains a
mixture of examples that require the knowledge and use of standard results and the
above techniques.

242

9 Differentiation 2 – Further Techniques

Exercise 8
Differentiate the following functions using the appropriate techniques and results.
1 f1x2 ⫽ x2 ⫺ 5x ⫹ 9

2 y ⫽ 12x ⫺ 72 3

3 f1x2 ⫽ cos 8x ⫺ 29x

4 y ⫽ sec x ⫺ e5x

5 f1x2 ⫽ x3e⫺4x

6 y ⫽ x2 ln x

7 f1x2 ⫽
10 y ⫽

sin 3x
ex

8 y⫽

log2 x
1x ⫺ 42 3

13 y ⫽ 6 sin⫺1 2x
16 y ⫽

ln1cot x2
ex

9 f1x2 ⫽ 3x sin x

11 f1x2 ⫽

x2 ln x
x⫹9

12 y ⫽ 3 cos 2x sin 4x

14 f1x2 ⫽

cos⫺1 x
3x2

15 y ⫽ x sin x ln x

17 Find f¿122 for f1x2 ⫽
1
x
tan⫺1¢ ≤.
x
4

18 Find f¿142 for f1x2 ⫽
20 Find

x sin x
e4x

19 Find

x3 1x ⫺ 72 2
12x ⫺ 12 3

.

dy
for x2y ⫹ exy2 ⫽ 9.
dx

dy
for x4y3 ⫺ y sin x ⫽ 2.
dx

9.9 Further differentiation problems
The techniques covered in this chapter can also be combined to solve differentiation
problems of various types, including equations of tangents and normals, and stationary
points. Problems of this type are given in Exercise 9.

Example
Find the stationary point for

y
⫽ x and determine its nature.
ex

In this case, it is easiest to consider this as y ⫽ xex.
dy
⫽ exx ⫹ ex
Using the product rule,
dx
⫽ ex 1x ⫹ 12
For stationary points,

dy
⫽0
dx

Hence ex 1x ⫹ 12 ⫽ 0
1 x ⫽ ⫺1
At x ⫽ ⫺1,

y ⫽ e⫺1 # ⫺1
1
⫽⫺
e

1
Hence the stationary point is ¢⫺1, ⫺ ≤
e
d2y
dx2

⫽ ex 1x ⫹ 12 ⫹ ex

⫽ ex 1x ⫹ 22

243

9 Differentiation 2 – Further Techniques

So, at x ⫽ ⫺1,

d2y

dx2
1

e

So

d2y
dx2

⫽ e⫺1 112

1
7 0, therefore ¢⫺1, ⫺ ≤ is a local minimum turning point.
e

Exercise 9
1 Find the gradient of the tangent to y ⫽ tan⫺1 3x where x ⫽

1 .
23

2 Find the gradient of the tangent to y ⫽ ln21 ⫺ cos 2x where x ⫽
3 Given y ⫽

p.
4

4x
, find the rate of change where x ⫽ 2.
e 1x ⫹ 22
x

4 Find the equation of the tangents to x2y ⫹ y2 ⫽ 6 at x ⫽ 1.
5 Find the gradient of the tangent to 2x ln x ⫺ y ln y ⫽ 2e11 ⫺ e2 at the point 1e, e2 2.
6 Find the value of

d2y

when x ⫽ p for

dx2

7 Find the stationary points of y ⫽

y
sin 2x .

x
ex

x2 .
ex

8 Find the stationary points of y ⫽ 4x2 ln x, x 7 0.
9 Find the stationary points, and their nature, of the curve given by y ⫽
10 Show that the gradient of the tangent to the curve given by
xy
⫹ sin x ln x ⫽ cos x ⫹ 1 at x ⫽ p is ln p.
p

Review exercise

✗ 1 Differentiate these functions.
M
C

7

4

1

M–

M+

CE

%

8

9

5

6

÷

2

3

+

0

ON
X

=

7

a y ⫽ 513x ⫺ 22 4

b f1x2 ⫽

d f1x2 ⫽ 6e8x

e y ⫽ ln 6x ⫺ 3x

213 ⫺ 2x2 2

✗ 2 Differentiate these functions.
M
C

7

4

1

M–

M+

CE

%

8

9

5

6

÷

2

3

+

0

ON
X

=

a y ⫽ e4x sin 3x

b y ⫽ ln1x sin x2

3x ⫹ 4
d y ⫽ ln¢

2x ⫺ 1

e2x cos 3x
e y ⫽ log10¢

1x ⫹ 42 2

dy

✗ 3 Find dx for:
dy
✗ 4 Find dx for x
M
C

7

4

1

M–

M+

CE

%

8

9

5

6

÷

2

3

+

0

a 4y2 ⫺ 3x2y ⫽ 5

ON
X

=

2

M

M–

M+

ON

C

CE

%

X

7

8

9

5

6

÷

2

3

4

1

+

0

=

2

2

sin x ⫺ exy ⫽ 7.

✗ 5 Differentiate y ⫽ 2 tan
M
C

7

4

1

0

244

M–

M+

CE

%

8

9

5

6

÷

2

3

+

c y ⫽ 6t ⫺ sec 3t

ON
X

=

⫺1

¢

1 ⫹ cos x
≤.
sin x

c f1x2 ⫽

e5x
2x ⫹ 4

b x3 ⫽ y ln x

x3
.
ex

9 Differentiation 2 – Further Techniques

M
C

M–

M+

CE

%

8

9

5

6

÷

2

3

7

4

1

ON
X

=

+

0

M
C

M–

M+

CE

%

8

9

5

6

÷

2

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7

4

1

ON
X

M
C

M–

M+

CE

%

8

9

5

6

÷

2

3

7

4

1

ON
X

=

+

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M
C

M–

M+

CE

%

8

9

5

6

÷

2

3

7

4

1

M
C

ON
X

=

+

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M–

M+

CE

%

8

9

5

6

÷

2

3

7
4
1

X

=

M

M–

M+

C

CE

%

7

8

9

4

5

6

÷

1

2

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M+

CE

%

8

9

5

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÷

2

3

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7
4
1

X

9 A curve has equation x3y2 ⫽ 8.
Find the equation of the normal to the curve at the point (2,1).
[IB May 03 P1 Q10]

=

e2x sin x
for 0 6 x 6 p.
x⫹1

12 Show that the point P12, ⫺22 lies on the circle with equation

ON
X

=

+

0

[IB May 01 P1 Q4]

8 A curve has equation xy3 ⫹ 2x2y ⫽ 3.
Find the equation of the tangent to this curve at the point (1, 1).
[IB May 02 P1 Q17]

11 Find the stationary points of y ⫽

ON

+

0

M

x ⫽ 1 and y 7 0.

10 Find the stationary points of y ⫽ x2 tan⫺1 x.

ON

+

0

7 Find the gradient of the tangent to 3x2 ⫹ 4y2 ⫽ 7 at the point where

=

+

0

1
6 Find the exact value of the gradient of the tangent to y ⫽
where
x sin x
p.
x⫽
4

1x ⫹ 22 2 ⫹ 1y ⫺ 22 2 ⫽ 32 and the parabola with equation y2 ⫽ 12 ⫺ 4x.
Also show that these curves share a common tangent at P, and state the

M
C

M–

M+

CE

%

8

9

5

6

÷

2

3

7

4

1

M
C

7

4

1

C
7
4
1
0

X

=

M–

M+

CE

%

8

9

5

6

÷

2

3

+

0

M

ON

+

0

ON
X

[IB Nov 04 P1 Q5]

=

M–

M+

CE

%

8

9

5

6

÷

2

3
+

equation of this tangent.
d2y
.
13 If y ⫽ ln12x ⫺ 12, find
dx2
14 Consider the function f1t2 ⫽ 3 sec 2t ⫹ 5t.

ON
X

=

a Find f¿1t2 .
b Find the exact values of
i f1p2
ii f¿1p2.

[IB Nov 03 P1 Q8]

15 Consider the equation 2xy2 ⫽ x2y ⫹ 3.
a Find y when x ⫽ 1 and y 6 0.
b Find

dy
when x ⫽ 1 and y 6 0.
dx

[IB Nov 03 P1 Q15]

245

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