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NATIONAL UNIVERSITY OF SINGAPORE
Department of Mathematics
MA 1505 Mathematics I
Tutorial 1 (Solution Notes)
1. Let f(x) = 6x and g(x) =
√|3 − x|. Find an expression for (g ◦ f)(x) − (f ◦ g)(x).
Solution
By the definition of function composition, we know that
(g ◦ f)(x) = g[f(x)],
(f ◦ g)(x) = f [g(x)],
then
(g ◦ f)(x) = g[6
x] =
√|3 − 6
x|,
(f ◦ g)(x) = f [√|3 − x|] =
6√|3 − x| .
Hence,
(g ◦ f)(x) − (f ◦ g)(x) =
√|3 − 6
x| − 6√|3 − x| .
That is the answer. �
Tips
If you are not familiar with function composition, check this web page:
http://en.wikipedia.org/wiki/Composite function.
1
2. Find the first derivatives of the following functions.
(a) y =ax + b
cx + d(b) y = sinn x cos(mx)
(c) y = ex2+x3(d) y = x3 − 4(x2 + e2 + ln 2)
(e) y =
(sin θ
cos θ − 1
)2
(f) y = t tan(2√
t) + 7
(g) r = sin(θ +√
θ + 1) (h) s =4
cos x+
1
tan xPreliminaries
This question is testing your some basic rules of first derivatives:
First of all, please notice that there are three kinds of notations for differentiation that are
widely used nowadays, and will appear in the lectures, tutorials, and tests of MA1505 and
MA1506 frequently:
• Leibniz’s Notation:dy
dx.
• Lagrange’s Notation:
y′(x).
• Newton’s Notation:
y(x).
It is no use to distinguish these notations, just keeping in mind that all of the three notations
are the same.
For more information, please refer to: http://en.wikipedia.org/wiki/Derivative.
(1) Quotient Rule
If the function is of the form f(x) =A(x)
B(x), then
f ′(x) =[A(x)]′ ∗ B(x) − A(x) ∗ [B(x)]′
B(x) ∗ B(x).
Abbreviation: (A
B
)′=
A′B − AB′
B2.
(2) Product Rule
If the function is of the form f(x) = A(x) ∗ B(x), then
f ′(x) = [A(x)]′ ∗ B(x) + A(x) ∗ [B(x)]′.
Abbreviation:
(AB)′ = A′B + AB′.
2
(3) Chain Rule
There are two ways to express this rule, they are equivalent, and both of them are very
useful. You can choose your favorite.
(1) (f ◦ g)′(x) = f ′(g(x))g′(x);
(2)dy
dx=
dy
du· du
dx.
(4) Constant Rule
If the function is a constant function, then its derivative is 0.
f = Constant =⇒ f ′ = 0.
You can check your lecture notes for more details.
All of them are both very important rule, and you need to be familiar with them.
Solution
(a) Qn: y =ax + b
cx + d,
Ans: Use Quotient Rule: (A
B
)′=
A′B − AB′
B2.
Then,
y′ =(ax + b)′(cx + d) − (ax + b)(cx + d)′
(cx + d)2=
a(cx + d) − c(ax + b)
(cx + d)2=
ad − bc
(cx + d)2.
Here, the first equality is using quotient rule, the second is using constant rule, and the
last gets by some basic algebraic operations on the numerator.
(b) Qn: y = sinn x cos mx ,
Ans: Use Product Rule:
(AB)′ = A′B + AB′,
we get:
y′ = [sinn x · cos mx]′ = [sinn x]′ · cos mx + sinn x · [cos mx]′
Then use Chain Rule:
(f ◦ g)′(x) = f ′(g(x))g′(x),
we know:
[sinn x]′ = n sinn−1 x,
[cos mx]′ = −m sinmx.
Hence,
y′ = n sinn−1 x cos x cos mx − m sinn x sin mx.
3
(c) Qn: y = ex2+x3
Ans: y′ = ex2+x3(2x + 3x2) (use chain rule)
(d) Qn: y = x3 − 4(x2 + e2 + ln 2) ,
Ans: y′ = 3x2 − 8x, (note that e2 and ln 2 are constants)
(e) Qn: y =
(sin θ
cos θ − 1
)2
Ans: −2 sin θ(cos θ − 1)−2 (use quotient and chain rule)
(f) Qn: y = t tan(2√
t) + 7
Ans:√
t sec2(2√
t) + tan(2√
t) (use product and chain rule)
(g) Qn: r = sin(θ +√
θ + 1)
Ans: 2√
θ+1+12√
θ+1cos(θ +
√θ + 1) (use chain rule)
(h) Qn: s =4
cos x+
1
tan xAns: 4 tan x sec x − csc2x (use quotient rule) �
4
3. Coffee is drained from a conical filter into a cylindrical coffeepot at the rate of 10 in3/min.
(a) How fast is the level in the pot rising when the coffee in the cone is 5 in. deep?
(b) How fast is the level in the cone falling then?
(Volume of cone: 13× base area × height)
Preliminaries
You need to know two things before doing this question:
(1) Suppose f(t) is a function with variable t, then the natural meaning of f ′(t) or we writeddtf(t) is the change rate of f at time t.
(2) To solve word problem, the most important thing is to find some relations among the
known values and unknown values. 1
How to find the equations? Let’s use this question as an example:
(1) What is known?
We only know: the rate of volume change dV/dt:
Note that the rate of volume change in the cone is equal the rate of volume change
in the pot.dVc
dt=
dVp
dt= 10 in3/min,
where Vc(t) be the volume of coffee in the cone at time t and Vp(t) be the volume
of coffee in the pot at time t.
(2) What do we want to know?
We want to know how fast the levels are changing, that is the rate of height change
dh/dt.
(3) What are the relationships?
The relations are from the formula of volume:
Vp = base area × hp, Vc =1
3× base area × hc.
1Next semester, when you take the module MA1506, you will find nearly all the tutorial questions are word
problems. So it would be very helpful if you practise from now on.
5
Solution
(a) Let Vp(t) be the volume of coffee in the pot at time t and hp(t) be the level of coffee in
the pot at time t.
You must be familiar with the formula:
Vp = base area × hp = (πr2) × hp = (π · 32) × hp = 9πhp.
Then operated
dton both sides of the formula, and we have:
dVp
dt= 9π
dhp
dt
⇒ 10 = 9πdhp
dt
⇒ dhp
dt=
10
9π
(b) Let Vc(t) be the volume of coffee in the cone at time t and hc(t) be the level of coffee in
the cone at time t.
You must be familiar with the formula:
Vc =1
3× base area × hc =
1
3(πr2)hc =
1
3π(
hc
2)2hc =
π
12h3
c .
Here we use the information from the picture: the base radius r of the cone is half that of
the height hc, which follows from the property of similar triangles.
Then operated
dton both sides of the formula, and we have:
dVc
dt=
π
12× (3 × h2
c) ×dhc
dt
⇒ 10 =π
12× (3 × 52) × dhc
dt
⇒ dhc
dt=
8
5π
�
6
4. For the following functions, find y′ and y′′.
(a) x2/3 + y2/3 = a2/3, 0 < x < a, 0 < y
(b) y = (sin x)sin x, 0 < x < π2
(c) x = a cos t, y = a sin t
Preliminaries
To solve Implicit Differentiation needs quite technical tricks, you must gather your own
experience while practising.
Generally speaking, we proceed as follows:
• First, differentiate the equation we are given, and get a new equation.
• Then solve the new equation, and substitute the original equation.
Question (a) is a typical example.
Solution
(a) Qn: x2/3 + y2/3 = a2/3. 0 < x < a, 0 < y.
Ans: Differentiating the equality we get (notice that the right-hand-side is a constant,
and don’t forget the part “dy
dx” in the new equation, which is due to chain rule)
2
3x− 1
3 +2
3y−
13dy
dx= 0.
Since 0 < x < a and 0 < y (why do we need such conditions? 2), we can solve outdy
dx
from the new equation, and substitute y =32
√a
23 − x
23 from the original equation:
dy
dx= −
23x− 1
3
23y−
13
= −y13
x13
= −
√a
23 − x
23
x13
= −
√a
23 − x
23√
x23
= −√
a23 − x
23
x23
= −√
(a
x)
23 − 1;
Then we can get:
d2y
dx2=
d
dx
(dy
dx
)=
d
dx
(−
√(a
x)
23 − 1
)= −1
2
1√(a
x)
23 − 1
· d
dx
((a
x)
23 − 1
)
= −1
2
1√(a
x)
23 − 1
·(−2
3a
23 x− 5
3
)=
a23
3x53
√(a
x)
23 − 1
=a
23
3x43
√a
23 − x
23
.
2If the range of x contains 0, then x− 13 is meaningless; similarly, if the range of y contains 0, then y− 1
3 is
meaningless; Meanwhile, y �= 0 implies x �= a by the original equation x23 + y
23 = a
23 , that is the range of x does not
contain a.
Further questions (but not deal with differentiation): can we change the conditions to one of followings:
• 0 < x < a, 0 < y < a (ANS: Yes);
• 0 < x, 0 < y < a (ANS: Yes);
• 0 < x < a (ANS: No);
• 0 < x < a, y < a (ANS: No);
7
Preliminaries
Question (b) is another important kind of function of the form y = [f(x)]g(x).
For the function of this type, we usually solve it like this:
• First, operate the logarithmic function “ln(·)” on y = [f(x)]g(x), and get:
ln y = g(x) · ln f(x);
Note: to make ln f(x) meaningful, all the values of f(x) must be positive.
• Then, differentiate the new equation
(ln y)′ = [g(x) · ln f(x)]′ =⇒ y′
y= [g(x) · ln f(x)]′;
• Finally, we can solve out y′ and y′′:
y′ = y · [g(x) · ln f(x)]′
= [f(x)]g(x) · [g(x) · ln f(x)]′,
y′′ = (y′)′
=[y · [g(x) · ln f(x)]′
]′= y′ · [g(x) · ln f(x)]′ + y · [g(x) · ln f(x)]′′
= [f(x)]g(x) · [[g(x) · ln f(x)]′]2
+ [f(x)]g(x) · [g(x) · ln f(x)]′′
You will see a lot of examples in such a form in the coming tutorials.
Solution
(b) Qn: y = (sin x)sin x, 0 < x < π2 .
Ans: Since 0 < x < π2 , we know sin x > 0, and we can apply the above scheme:
ln y = sinx ln sin x,
=⇒ y′
y= cos x ln sinx + cos x,
=⇒ y′ = y(1 + ln sin x) cos x,
=⇒ y′′ = y′(1 + ln sinx) cos x + y
[(1 + ln sinx)(− sin x) +
cos2 x
sin x
]
= y(1 + ln sin x)2 cos2 x + y
[cos2 x
sin x− (1 + ln sin x) sin x
].
Here, when we calculate y′′, we don’t substitute the value of y into y′, which makes the
whole calculation more concise.
Hence, we substitute y, and get:
y′ = y(1 + ln sinx) cos x
= (sin x)sin x(1 + ln sinx) cos x,
y′′ = y(1 + ln sinx)2 cos2 x + y
[cos2 x
sin x− (1 + ln sin x) sin x
]
= (sin x)sin x
[(1 + ln sinx)2 cos2 x +
cos2 x
sin x− (1 + ln sin x) sin x
].
8
Preliminaries
Question (c) is concerning parameter equations.
For these question, the solution scheme is:
• First, treat the parameter equations as many dependent equations, and for each
equation, differentiating on both sides;
• Then, try to combine the new equations together to get your target. Here, com-
bination means by using four arithmetic operations (+,−,×,÷), and usually is by
using division ÷.
Solution
(c) Qn: x = a cos t, y = a sin t.
Ans:
Differentiate the parameter equations separately:
x = a cos t =⇒ dx
dt= a
d
dtcos t = −a sin t
y = a sin t =⇒ dy
dt= a
d
dtsin t = a cos t
Our target is y′ (that is,dy
dx), then you can easily find the solution:
dy
dx=
dy
dtdx
dt
=a cos t
−a sin t= − cot t,
d2y
dx2=
d
dx(dy
dx) =
d(dydx )
dx=
ddt(
dydx)
dxdt
=
d
dt(− cot t)
dx
dt
=
1
sin2 t−a sin t
= − 1
a sin3 t.
�
Remark
i. For the second derivative, we can also do like this: consider t as a function t(x),
actually, t = cos−1 xa , thus, y′ = − cot t = − cot(cos−1 x
a ). Using Chain rule, we
know that
y′′ = [− cot(cos−1 x
a)]′ = [− cot′(cos−1 x
a)] · [cos−1 x
a]′ = [− cot t]′ · dt
dx=
[− cot t]′dxdt
,
which can get the same answer.
ii. When you have practised enough exercises, and are familiar with these questions,
you can directly write down the last two rows as the official solution does.
9
5. For each of the following functions:
(a) y =x + 1
x2 + 1, x ∈ [−3, 3] (b) y = (x − 1)
3√
x2, x ∈ (−∞,∞).
determine:
(i) the critical points;
(ii) the intervals where it is increasing and decreasing;
(iii) the local and absolute extreme values.
Preliminaries
(i) Critical Points:
• y′(x) = 0 or y′(x) fails to exist =⇒ x is a critical point;
• Critical points are all interior points.
(ii) The Increasing Intervals and the Decreasing Intervals:
• y′(x) > 0 for all interior points x in the interval I =⇒ I is an increasing interval;
• y′(x) < 0 for all interior points x in the interval I =⇒ I is an decreasing interval.
(iii) The Local Extreme Values and Absolute Extreme Values:
We use the following graph as an example:
The picture can be titled as “Among the blind the one-eyed man is king”.
Here, we have 9 critical points with 2 end points, and their values are called extreme
values.
• The values of the points A,C,E,G, I,K are called local minimal values, where the
value of G is the absolute minimal values; 3
• The values of the points B,D,F,H, J are called local maximal values, where the
value of H is the absolute maximal value.
3Please notice that the absolute extreme values are also counted as local extreme values.
10
(iv) Given a function, how to draw the graph of the function?
Generally speaking, to draw the graph of a given function, here are the standard steps:
(a) Find out all the critical points (these points divide the whole interval into several
small sub-intervals);
(b) For each critical point x, find the value of f(x);
(c) Draw all the critical points with their values, and connect all the points smoothly
except for the point that does not have first-order derivatives.
Then you draw a draft of the graph, and this is also a INFORMAL way to determine
whether an interval is increasing or decreasing;
Solution
(a) y =x + 1
x2 + 1, x ∈ [−3, 3].
Ans: Using quotient rule, we know that
y′ =(x + 1)′ · (x2 + 1) − (x + 1) · (x2 + 1)′
(x2 + 1)2=
(x2 + 1) − (x + 1) · (2x)
(x2 + 1)2=
−x2 − 2x + 1
(x2 + 1)2,
and then let y′ = 0, we can get the critical points are x = −1 ±√2.
So critical points are x = −1 ±√2 and endpoints are x = ±3.
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
y′ < 0 if − 3 ≤ x < −1 −√2,
y′ = 0 if x = −1 −√2,
y′ > 0 if − 1 −√2 < x < −1 +
√2,
y′ = 0 if x = −1 +√
2,
y′ < 0 if − 1 +√
2 < x ≤ 3.
Hence y is decreasing in [−3,−1 −√2), increasing in (−1−√
2,−1 +√
2), and decreasing in
(−1 +√
2, 3].
Trick: To determine whether y′ > 0 or y′ < 0 in some interval, you have two short-
cuts:
• Pick up any interior point s in the interval, and see whether y′(s) is positive or
negative. (Note that y′(s) can not be 0. Why? 4)
– y′(s) > 0 =⇒ y′ > 0 =⇒ the interval is increasing.
– y′(s) < 0 =⇒ y′ < 0 =⇒ the interval is decreasing.
• Find out the endpoints a and b of the interval (a < b), and compare the values
of y(a) and y(b).
– y(a) < y(b) =⇒ y′ > 0 =⇒ the interval is increasing.
– y(a) > y(b) =⇒ y′ < 0 =⇒ the interval is decreasing.
4Since all the points satisfying y′(x) = 0 have been pick out as critical points.
11
So local minima is:
y(−1 −√
2) = − 1
2(√
2 + 1), y(3) =
2
5
and local maxima is:
y(−1 +√
2) =1
2(√
2 − 1), y(−3) = −1
5.
Since
− 1
2(√
2 + 1)< −1
5<
2
5<
1
2(√
2 − 1),
so absolute minima is minx∈[−3,3]
y = − 1
2(√
2 + 1)at x = −1 −√
2
and absolute maxima is maxx∈[−3,3]
y =1
2(√
2 − 1)at x = −1 +
√2.
(b) y = (x − 1)3√
x2, x ∈ (−∞,∞).
Ans:
y′ =[
(x − 1)3√
x2]′
=[
(x − 1) · x 23
]′= [ (x − 1) ]′ · x 2
3 + (x − 1) ·[
x23
]′= x
23 +
2
3(x − 1)x− 1
3
=5x − 2
3x1/3(x = 0)
and then let y′ = 0, we can get the critical points are x =2
5.
Note that y′ does not exist at x = 0.
So the critical points are x = 0 and x =2
5.
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
y′ > 0 if x < 0,
y′ does not exist if x = 0,
y′ < 0 if 0 < x < 25 ,
y′ = 0 if x = 25 ,
y′ > 0 if x > 25 .
Hence y is increasing in (−∞, 0), decreasing in (0,2
5), and increasing in (
2
5,∞).
So local maxima is y(0) = 0, and local minima is y(2
5) = −3
5(2
5)
23 .
Since limx→−∞ y = −∞, lim
x→∞ y = ∞, so there is no absolute extremes.5 �
5There is a shortcut to find absolute extremes, that is using Preliminary (iv). You can find more details in the
tutorial webcast folder:
http://cid-957f4598794dd719.skydrive.live.com/self.aspx/Tutorial Webcast.
12
6. Ornithologists have determined that some species of birds tend to avoid flights over large
bodies of water during daylight hours. It is believed that more energy is required to fly over
water than land because air generally rises over land and falls over water during the day. A
bird with these tendencies is released from an island that is 5 km from the nearest point B on
a straight shoreline, flies to a point C on the shoreline, and then flies along the shoreline to
its nesting area D. Assume that the bird instinctively chooses a path that will minimize its
energy expenditure. Points B and D are 13 km apart. If it takes 1.4 times as much energy
to fly over water as land, find the distance between B and C.
Preliminaries
This is an application of the maxima-minima problem (Question 5). The difficult is to un-
derstand the question, and transform the sentences into mathematical formulas.
What we know can be described as the graph.
Solution
Let x be the distance between B and C. Suppose the energy that it takes to fly over land is
1 unit per km, then it will take 1.4 unit per km to fly over water.
The total energy is given by the function
f(x) = 1.4√
52 + x2 + (13 − x).
Then
f ′(x) =1.4x√52 + x2
− 1,
f ′(x) = 0 =⇒ 1.4x =√
52 + x2 =⇒ 1.96x2 = 52 + x2 =⇒ x = 5.103,
and use the scheme of Question 5:
f ′(x) < 0, when x < 5.103, and f ′(x) > 0, when x > 5.103,
which shows that the point x = 5.103 is an absolute minimum. �
13
Practice Exercises I:Here are some questions appeared in past years’ tutorials. I attach them here, just for reference:
Pra 1.1 Find the first derivatives of the following functions.
(a) y =1
x+
1
x2+
1
x3. (b) y =
ax + b
cx + d. (c) y = sinn x cos(mx) .
(d) y = tan4 x . (e) y = ex2+x3. (f) y = arcsin
1
x, |x| > 1.
(g) y = x3 − 4(x2 + e2 + ln 2) . (h) y =
(sin θ
cos θ − 1
)2
.
(i) y = t tan(2√
t) + 7 . (j) r = sin(θ +√
θ + 1) .
(k) s =4
cos x+
1
tan x. (l) y = (3 + cos3 3x)−1/3.
Pra 1.2 Determine whether the function f(x) = x2|x| is differentiable at x = 0. If so, find f ′(0) .
Pra 1.3 Find the derivative of the following functions at the given points.
(a) y = anxn + an−1xn−1 + · · · + a1x + a0, find y′(0), y′(1) .
(b) y = (x − 1)(x − 2)2(x − 3)3, find y′(1), y′(2), y′(3) .
(c) y = sin x + cos x, find y′(0), y′(π4 ), y′(π
2 ) .
Pra 1.4 Show that the tangents to the curve y =π sinx
xat x = π and x = −π intersect at right
angle.
Pra 1.5 For each of the following functions, determine:
(i) the critical points;
(ii) the intervals where it is increasing and decreasing;
(iii) the local and absolute extreme values.
Qn (a). y = x − ln(1 + x), x ∈ (−1,∞).
Ans:
y′ = 1 − 1
|1 + x|=⇒ y′ = 1 − 1
1 + x=
x
1 + x(∵ 1 + x > 0, when x > −1)
=⇒ The critical point is x = 0.
=⇒ y′ < 0 if − 1 < x < 0 and y′ > 0 if 0 < x < ∞.
Thus, y is decreasing in (−1, 0) and increasing in (0,∞).
The local minima is y(0) = 0, and the unique local minima is also the absolute minima.
Since limx→−1+
y = ∞ and limx→∞ y = ∞, there is no local maxima, and so no absolute maxima.
Qn (b). y = 2 tan x − tan2 x, x ∈ [0, π2 ).
14
Ans:
y′ =2
cos2 x− 2 tan x · 1
cos2 x=
2
cos2 x(1 − tan x).
Tips: If you can not remember the derivative of tan x, you can write tan x =sin x
cos x,
and then use quotient rule.
Suggestion: Remember it since it is frequently used, and it helps you save a lot of
time in the exam.
So the critical point is x = π4 , and y is increasing in [0, π
4 ) and decreasing in (π4 , π
2 ).
Thus, local minima is y(0) = 0, and local maxima is y(π4 ) = 1.
Since limx→π2− y = −∞, absolute maxima is max
0≤x< π2
y = 1 at x = π4 and there is no absolute
minima. �
Remark: Qn (b) tells us that if there is a unique local minima, which is a candidate
for absolute minima, we can not say that this local minima is absolute minima. The
case is similar for local maxima.
Qn (c). y = sin3 x + cos3 x, x ∈ (−∞,∞)
Pra 1.6 What value of θ will maximize the area of the trapezium as shown?
Ans. π6
Pra 1.7 Given a constant c > 2, prove that the equation x3 −3x+ c = 0 has exactly one real root.
Practice Exercises II:Besides these, you can try the following exercises:
(1) There are plenty of exercises in your text book, Thomas’ Calculus, for this tutorial, you
can refer to the exercises in Chapter 1, 3, and 4, which are available in the shared folder.
The solutions to these exercises can be found in the appendix.
(2) If you do not like the style of Thomas’ Calculus, you can try James’ Calculus, Chapter
1–4 (belonging to intermediate level of Calculus), which is a very popular textbook on
Calculus as well.
(3) The workbooks 11 and 12 from HELM (Help Engineer Learn Mathematics) project
(belonging to advanced level of Calculus) is also a very wonderful material for you to practise.
Solutions to these exercises are available in the shared folder.
Last but most important: Choose ONE of the three books to practise is enough!
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