Review ()
√
A sample test
Question . — Let f (x) = x x + 1.
a. Use the definition of the derivative to compute f 0 (x).
b. Write an equation of the line tangent to the graph of f at the point with x coordinate 2.
Question . — Let f be the function defined by


p

x cos(π/x)
f (x) = 

0
Another sample test
x
Question . — Given f (x) =
.
2x + 1
a. Use the definition of the derivative to find f 0 (x).
b. Find all points on the graph of f at which the tangent line has y intercept 2.
Question . — a. Use the definition of the logarithm in terms of area to show that
if x > 0, and
if x 6 0.
Find all values of p for which f is differentiable at 0, but f 0 is not continuous at 0. For full credit,
write a complete and well-organized solution.
Question . — Show that if f and g are rational functions whose graphs are tangent where x = a,
then f (x) − g(x) = (x − a)2 h(x), where h is a rational function whose domain includes a.
Question . — Compute each indicated derivative. For full credit, follow all instructions and
use the variable and function names as given.
√
dy
x(x3 − 5)3 (5x6 + 1)2
dx
, where y = x x + 1 −
, where xy = y x .
.
b. Find
dx
dy
(4x7 + 20)2
√
w
dw
c. Given f (x) = cot 2x + log7 (3x + sec 5x) , find f 0 (x).
d. Find
, if
= ln(rw).
dr
tan(r) + 1
dz e. Find and simplify
, where z = log
ecos 3y + tan( 23 y) .
(sin y)
dy 1
a. Find
y= 6 π
Question . — You are given the curve defined by y 3 + 3xy = x3 .
dy
and simplify your answer.
dx
6
b. Find an equation of the line tangent to the curve at the point 12
7 ,7 .
a. Find
c. At what point in the second quadrant is the tangent line horizontal?
d. Find all points on the curve at which the tangent line has slope 1.
Question . — The tangent line to the graph of a function f at the point where x = 1 is defined
by 2x + 3y = 5. Also, f 00 (1) = 7 and f 000 (1) = −9. Let g(x) = f 0 xf (x) and h(x) = f e2x .
a. Find an equation of the line tangent to the graph of g where x = 1.
b. Find an equation of the line tangent to the graph of h where x = 0.
c. Find g 00 (1).
Question . — For each equation, find all points on its graph where the tangent line is horizontal.
a. y = sin(2x) − 2 sin x
Winter 
Calculus I (-nya-/)
2
b. y = xx
Question . — A circle has centre (0, 0) and radius f > 0. A parabola has vertex (0, 0) and focus
(0, f ). Find all lines tangent to both curves.
[ This is due to Rubén Calzadilla-Badra. ]
Question . — Determine a formula defining f (n) (x), where f (x) = x2 e−3x .
Question . — State and prove the product rule for differentiation.
d
1
(ln x) =
dx
x
if x > 0.
x n
b. Using properties of the logarithm and of real exponents, compute lim 1 +
.
n→∞
n
Question . — Show that if the functions f , g are differentiable at a, f (a) = g(a), and f (x) < g(x)
for x , a, then f 0 (a) = g 0 (a).
Question . — Find the indicated derivative. For full credit use the variable names and notation
as given. You may write only the answers, and you need not simplify them.


 √

3
p
dp
4
du

π
5
a. Find
if u =  r − √
+ e  sin(πe ).
if p = tan q3 sin(q) + cos(ln q) .
b. Find
5
dr
dq
r3
sec
x
+
log2 (cos x)
dx
c. Find f 0 (x) if f (x) = √
.
d. Find
if y sin(x2 y) = 1 − cos(2x + 3y).
3
2
dy
x −x +7
dy
e. Find
if y = tan (sin x)cot x .
dx
Question . — You are given the curve defined by (x2 + y 2 )2 − 12(x2 + y 2 ) = 36x2 .
dy
a. Find and simplify
.
dx
b. Find equations of all tangents to the curve where it meets the line defined by y = x.
c. Find all points at which the line tangent to the curve is horizontal.
d. Find all points at which the line tangent to the curve is vertical.
Question . — Find all real numbers α, β and γ such that if
d2 y
= 8(1 + αy)(1 + βy).
dx2
Question . — Find all values of x at which the tangent to the graph of f is horizontal.
a. f (x) = x(ln x)2
b. f (x) = (x − 2)2 (x2 + 2x − 12)3
y = tan2 (γx),
then
Question . — The position of a particle along the x axis is described by the equation
x = (2t − 1)3 (t − 2)6 ,
for
t > 0,
where t is measured in seconds and x is measured in metres.
a. When is the particle at rest?
b. When is the particle moving to the left?
c. What distance does the particle travel during the first three seconds?
Question . — Find the monic poynomial of degree four, y = a + bx + cx2 + dx3 + x4 , whose graph
is tangent to the line y = −3x − 2 where x = −1, and is tangent to the line y = 6 − 7x where x = 1.
Question . — State and prove the quotient rule for differentiation.
Review () — Solutions to ‘A sample test’
Solutions to ‘A sample test’
√
Solution to question . — a. If f (x) = x x + 1, then
√
√
t 2 (t + 1) − x2 (x + 1)
f (t) − f (x) = t t + 1 − x x + 1 = √
,
√
t t+1+x x+1
where the numerator is t 3 − x3 + t 2 − x2 = (t − x)(t 2 + tx + x2 + t + x). Therefore,
f (t) − f (x)
3x2 + 2x
3x + 2
t 2 + tx + x2 + t + x
= √
= √
.
= lim √
√
t→x
t→x t t + 1 + x x + 1
t−x
2x x + 1 2 x + 1
√
8 = 4 √3, it follows that the tangent line is defined by
b. Since f (2) = 2 3 and f 0 (2) = 2√
3
3
f 0 (x) = lim
√
√
y − 2 3 = 43 3(x − 2),
or equivalently
√
4x − 3 y = 2.
Solution to question . — The function f is differentiable at 0 if, and only if,
f (x)
lim
= 0,
x→0+ x
or equivalently
if
x > 0,
and hence
f 0 (x) = pxp−1 cos(π/x) + πxp−2 sin(π/x).
By the same reasoning as above, f 0 (x) → 0 as x → 0+ if, and only if, p > 2.
Therefore, f is differentiable at 0 but f 0 is not continuous at 0 if, and only if, 1 < p 6 2.
Solution to question . — By combining terms, f (x) − g(x) = N (x)/D(x) for polynomials N (x)
and D(x) with D(a) , 0. The graphs of f and g are tangent where x = a, so f (a) = g(a) and
f 0 (a) = g 0 (a). Then N (a) = 0, so division gives N (x) = (x − a) ((x − a)q(x) + r), where q(x) is a
polynomial and r is a real number. Since f (a) − g(a) = 0 and f 0 (a) − g 0 (a) = 0, the definition of the
derivative gives
x→a
f (x) − g(x)
(x − a)q(x) + r
N (x)
r
= lim
= lim
=
,
x→a (x − a)D(x) x→a
x−a
D(x)
D(a)
which implies that r = 0, and therefore, f (x) − g(x) = (x − a)2 q(x)/D(x), as required.
Solution to question . — a. Write y = u − w, so that logarithmic differentiation gives
)
(
) p
(
log(x2 + 1)
du p
d log(x2 + 1)
2x
x
x
−
= x2 + 1
= x2 + 1
dx
dx
x
x(x2 + 1)
x2
n
o
−2
2
1/x−1
2
2
2
= x (x + 1)
2x − (x + 1) log(x + 1) ,
and thus
2
e. If
log ecos 3y + tan( 32 y)
cos
3y
3
z = log
e
+ tan( 2 y) =
,
(sin y)
log(sin y)
o
dw
d n
=w
log|x| + 3 log|x3 − 5| + 2 log|5x6 + 1| − 2 log|4x7 + 20|
dx
dx
(
)
x(x3 − 5)3 (5x6 + 1)2 1
9x2
60x5
56x6
=
+
+
−
,
x x3 − 5 5x6 + 1 4x7 + 20
(4x7 + 20)2
dy
du dw
=
−
.
dx
dx
dx
then
cos 3y sin 3y + 3 sec2 ( 3 y) cos(y) log ecos 3y + tan( 3 y)
dz −3e
2
2
2
= −
,
2
dy
ecos 3y + tan( 32 y) log(sin y)
sin(y) log(sin y)
lim xp−1 cos(π/x) = 0,
x→0+
since ±xp−1 → 0 as x → 0+ . So f is differentiable at 0 if, and only if, p > 1, in which case f 0 (0) = 0.
Next, if x < 0 then f 0 (x) = 0, and if x > 0 then
0 = lim
b. Writing the equation as y log x − x log y = 0 and then differentiating implicitly with respect
to y gives
log x − x/y x(x − y log x)
dx
=−
=
.
dy
y/x − log y y(y − x log y)
c. The chain rule gives
)
(
3 + 5 sec 5x tan 5x
.
f 0 (x) = − csc2 2x + log7 (3x + sec 5x) 2 +
(3x + sec 5x) log(7)
√
d. Writing the equation as (1 + tan r) log(rw) − w = 0 and differentiating implicitly with
respect to r gives
2
log(rw) sec2 r + (1 + tan r)/r 2w r log(rw) sec r + tan r + 1
dw
√
=−
=
.
dr
r ( w − 2(1 + tan r))
(1 + tan r)/w − 1 w−1/2
lim xp−1 cos(π/x) = 0.
x→0+
This is impossible if p 6 1, since then f (1/n)/(1/n) = (−1)n n1−p , which is alternately positive and
negative and has absolute value > 1 if n is a positive integer. On the other hand, if p > 1 then
−xp−1 6 xp−1 cos(π/x) 6 xp−1
Winter 
Calculus I (-nya-/)
and so
dz dy y= 16 π
=
√
1 √3 log(1 + 1)
3
2
−
=
−
.
log 2
1 log( 1 ) 2
(1 + 1) log( 12 )
2
2
−3 + 3
Solution to question . — a. Writing the given equation as y 3 − x3 + 3xy = 0 and differentiating
implicitly with respect to x gives
dy
−3x2 + 3y x2 − y
=−
= 2
.
dx
3y 2 + 3x
y +x
6
b. The slope of the tangent line to the curve at the point 12
7 , 7 is
2
12 − 6
dy 24 − 7 17
7
7
12
= 2
=
= 20 ,
and so
y − 76 = 17
20 x − 7 ,
dx
6
+
14
6
12
12
6
+
x= 7 , y= 7
7
7
or 17x − 20y = 12, defines the tangent line in question.
dy
c. The tangent line to the graph is horizontal where dx = 0, i.e., where
x2 − y
= 0,
y2 + x
or equivalently,
x2 − y = 0
and
y 2 + x , 0.
Replacing y by x2 in the given equation yields
x6 + 3x3 = x3 ,
x3 (x3 + 2) = 0.
√
√
√
Apart from the origin, this gives x3 = −2, or x = − 3 2, and y = (− 3 2)2 = 3 4. So the tangent line to
√
3 √
3
3
3
the graph of y + 6xy = x is horizontal at the point (− 2, 4). in the second quadrant.
d. The tangent line to the curve has slope 1 where
dy
= 1,
dx
i.e.,
x2 − y = y 2 + x
or
or
0 = x2 − y 2 − (x + y) = (x + y)(x − y − 1).
Write the equation of the curve (x − y)(x2 + xy + y 2 ) − 3xy = 0; if x − y − 1 = 0 on the curve, then
0 = x2 − 2xy + y 2 = (x − y)2 , or x − y = 0, which is impossible. So there are no points on the curve
Review () — Solutions to ‘A sample test’
where x − y − 1 = 0. So the tangent line to the curve has slope 1 where x + y = 0 and xy , 0.
Replacing y by −x in the equation of the curve gives
−x3 − 3x2 = x3 ,
0 = 2x3 + 3x2 = x2 (2x + 3),
or
so x = − 23 and y = 32 . Therefore, the tangent line to the curve has slope 1 at the point − 32 , 23 .
Solution to question . — From the given equation of the tangent line to the graph of f where
x = 1, and the other given information, it follows that
f (1) = 1,
a. Since g(x) = f 0 (xf (x)),
f 0 (1) = − 23 ,
g(1) = f 0 (f (1)) = f 0 (1) = − 23 ,
so
f 00 (1) = 7
and
and
f 000 (1) = −9.
g 0 (x) = f 00 (xf (x)) f (x) + xf 0 (x) ,
g 0 (1) = f 00 (f (1)) f (1) + f 0 (1) = f 00 (1) 1 − 32 = 73 .
Therefore, 7x − 3y = 9
is an
equation of the line tangent to the graph of g at the point where x = 1.
b. Since h(x) = f e2x ,
h(0) = f (e2·0 ) = f (1) = 1,
and
h0 (x) = f 0 (e2x ) · e2x · 2,
so
h0 (1) = f 0 (1) · 2 = − 43 .
Therefore, 4x + 3y = 3 is an equation of the tangent line to the graph of h at the point where x = 0.
c. Since g 0 (x) = f 00 (xf (x)) (f (x) + xf 0 (x)), it follows that
2
g 00 (x) = f 000 (xf (x)) f (x) + xf 0 (x) + f 00 (xf (x)) 2f 0 (x) + xf 00 (x) ,
and so
Winter 
Calculus I (-nya-/)
2
116
g 00 (1) = f 000 (1) 1 − 23 + f 00 (1) − 34 + 7 = −9 · 19 + 7 · 17
3 = 3 .
dy
= 2 cos(2x) − 2 cos x = 2(2 cos2 x − cos x − 1) = 2(2 cos x + 1)(cos x − 1),
dx
which is equal to zero if, and only if, either cos x = − 12 , i.e., x is ± 23 π + 2kπ, or else cos x = 1, i.e.,
√
√
√
x is 2kπ, where in each case k is an integer. If x = 32 π + 2kπ then y = − 12 3 − 3 = − 23 3, if
√
√
√
2
1
3
x = − 3 π + 2kπ then y = 2 3 + 3 = 2 3 and if x = 2kπ then y = 0. Therefore, the tangent line
√ to the curve is horizontal at the points ± 23 π + 2kπ, ∓ 3 and (2kπ, 0), where k is an integer (and
at no other points).
2
b. If y = xx , then logarithmic differentiation gives
2 d 2
2
2
dy
= xx
x log x = xx (2x log x + x) = xx +1 (2 log x + 1),
dx
dx
−1
= e−1/(2e) .
Solution to question . — The parabola is defined by x2 = 4f y and the circle is defined by
x2 + y 2 = f 2 . A line is tangent to the parabola at (x, y) and to the circle at (p, q) if, and only if,
y −q
p
x
x
x2 = 4f y,
p2 + q2 = f 2 ,
=
and
=− .
2f
x−p
2f
q
The third and first equations give 2x2 − 2px = 4f y − 4f q = x2 − 4f q, and the second equation gives
(x − p)2 = p2 − 4f q = f 2 − 4f q − q2 ,
or
p2
(x − p)2 = 2 (2f + q)2 ,
q
√
Therefore, 2f + q = −q 5, or
q=
−2f
f
√ =− ,
τ
1+ 5
and
and hence
p2 = f 2 − q2 =
(2f + q)2 = 5q2 .
f2
,
τ
so
f
p = ±√ ,
τ
where τ is the positive solution of x2 = x + 1 (the so-called golden section). Thus, the tangent
√
√
lines have slope ± τ, are defined by y + f τ = ±x τ, and make contact with the parabola at the
√
points (±2f τ, f τ).
Solution to question . — First observe that if y = e−3x g(x), then
dy
= e−3x (−3)g(x) + e−3x g 0 (x) = e−3x −3g(x) + g 0 (x) .
dx
Using this to compute and then carefully simplify the first four derivatives of f (x) = e−3x x2 , gives
f 0 (x) = e−3x (−3x2 + 2x),
f 00 (x) = e−3x −3(−3x2 + 2x) − 6x = e−3x (9x2 − 6 · 2x + 2),
f (3) (x) = e−3x −3(9x2 − 6 · 2x + 2) + 18x − 6 · 2 = e−3x (−3)(9x2 − 6 · 3x + 3 · 2)
and
(4)
−3x
2
−3x
2
2
f (x) = e
(−3) −3(9x − 6 · 3x + 3 · 2) + 18x − 6 · 3 = e
(−3) (9x − 6 · 4x + 4 · 3).
This suggests the pattern
f (n) (x) = e−3x (−3)n−2 9x2 − 6nx + n(n − 1) ,
which is readily verified in the four cases displayed. Since
o
d n −3x
e
(−3)n−2 9x2 − 6nx + n(n − 1) = e−3x (−3)n−2 −3(9x2 − 6nx + n(n − 1)) + 2 · 9x − 6n
dx
= e−3x (−3)n−1 9x2 − 6(n + 1)x + (n + 1)n ,
Solution to question . — a. If y = sin(2x) − 2 sin x, then
which is equal to zero if, and only if, log x = − 12 , or x = e−1/2 , where y = (e−1/2 )e
Thus, the curve has one horizontal tangent line, at the point e−1/2 , e−1/(2e) .
The fourth equation implies that
(x − p)2 + (2f + q)2 = 5f 2 .
it follows that the pattern will continue for all higher derivatives.
Solution to question . — Where u and v are differentiable functions of x, so is y = uv, and
dy
du
dv
=
v+u .
dx
dx
dx
Proof. — The proof is a direct calculation apart from the last equation, where the continuity of
differentiable functions insures that v 0 → v as x0 → x.
0
dy
y0 − y
u −u 0
u 0 v 0 − uv 0 + uv 0 − uv
v0 − v
= lim 0
= lim
= lim
·v +u · 0
0
0
0
0
0
dx x →x x − x x →x
x −x
x −x
x →x x − x
du
dv
=
v+u .
dx
dx
Review () — Solutions to ‘Another sample test’
Solutions to ‘Another sample test’
Solution to question . — a. If f (x) = x/(2x + 1), then
f (x0 ) − f (x) =
x0 (2x + 1) − (2x0 + 1)x
x0
x
x0 − x
−
=
=
,
0
0
0
2x + 1 2x + 1
(2x + 1)(2x + 1)
(2x + 1)(2x + 1)
and so
f 0 (x) = lim
x0 →x
f (x0 ) − f (x)
1
1
.
= lim
=
x0 − x
(2x + 1)2
x0 →x (2x0 + 1)(2x + 1)
b. The tangent line to the graph of f has y intercept 2 where
x
−2
1
2x + 1
=
,
x−0
(2x + 1)2
or
x = (2x + 1)(x − 2(2x + 1)) = −(2x + 1)(3x + 2).
Solution to question . — a. If


3
 p

4
π

5

u =  r − √
+ e  sin(πe ) = r 5/3 − 4r −3/5 + eπ sin(πe ),
5
r3
then
du 5 2/3 12 −8/5 = 3r
+ 5r
sin(πe ).
dr
p
b. If p = tan q3 sin(q) + cos(ln q) , then



p
dp
sin(ln q) 
= sec2 q3 sin(q) + cos(ln q) 3q2 sin(q) + q3 cos(q) − p
 .
dq
2q cos(ln q)
c. If f (x) =
Equivalently, 0 = 6x2 + 8x + 2 = 2(x + 1)(3x + 1). Since f (−1) = 1 and f (− 13 ) = −1, it follows that
the tangent to the graph of f at the points (−1, 1) and (− 13 , −1) have y intercept 2 (and there are
no other such points on the graph of f ).
Solution to question . — a. If 0 < α < β then ln β − ln α is the area of the region defined by
α 6 x 6 β and 0 6 y 6 1/x, and so
β −α
β −α
< ln β − ln α <
,
β
α
1 ln β − ln α 1
<
< .
β
β −α
α
or
Therefore,
lim
α→β −
ln β − ln α 1
=
β−α
β
and
lim
β→α +
so by the definition of the derivative,
d
1
(ln x) = ,
dx
x
provided
ln β − ln α
1
= ,
β−α
α
x
x n
lim 1 +
= lim (1 + t)1/t = ex ,
n
t→0
x > 0.
where
x = nt.
Solution to question . — If f (x) < g(x) for x , a and f (a) = g(a), then f (x) − f (a) < g(x) − g(a)
for x , a. Therefore,
f (x) − f (a) g(x) − g(a)
>
x−a
x−a
if x < a
and
Since f and g are differentiable at a it follows that
f (x) − g(a) g(x) − g(a)
<
x−a
x−a
f 0 (a) = lim−
x→a
f (x) − f (a)
g(x) − g(a)
> lim−
= g 0 (a),
x→a
x−a
x−a
f 0 (a) = lim
g(x) − g(a)
f (x) − f (a)
6 lim
= g 0 (a),
x−a
x−a
x→a+
and
x→a+
which implies that f 0 (a) = g 0 (a).
y sin(x2 y) = 1 − cos(2x + 3y),
i.e.
y sin(x2 y) + cos(2x + 3y) = 1,
then implicit differentiation gives
o
dy
dn
= sec2 (sin x)cot x (sin x)cot x
cot(x) ln(sin x)
dx
dx
= sec2 (sin x)cot x (sin x)cot x − csc2(x) ln(sin x) + cot2(x) .
x=1
n→∞
d. If
sec x + log2 (cos x)
then, since log2 = (ln 2)−1 ln,
√
x3 − x2 + 7
sec x + log2 (cos x) (3x2 − 2x)
sec x tan x − (ln 2)−1 tan x
f 0 (x) =
.
−
√
2(x3 − x2 + 7)3/2
x3 − x2 + 7
sin(x2 y) + x2 y cos(x2 y) − 3 sin(2x + 3y)
dx
=−
.
dy
2xy 2 cos(x2 y) − 2 sin(2x + 3y)
e. If y = tan (sin x)cot x , then the chain rule and logarithmic differentiation give
b. Since the logarithm is one-to-one and continuous (by Part a, since it is differentiable), and
ln(1 + t)
d
= 1,
it follows that
lim (1 + t)1/t = e,
lim
=
(ln x) t
dx
t→0
t→0
and therefore
Winter 
Calculus I (-nya-/)
if x > a.
Solution to question . — Notice that the given equation can be written as r 4 − 12r 2 − 36x2 = 0,
where r 2 = x2 + y 2 is the square of the distance between (x, y) and the origin, and that this implies
that 12 6 r 2 6 48 for any point on the curve besides the origin (so the origin is an isolated point).
a. Differentiating implicitly with respect to x gives
dy
4xr 2 − 24x − 72x x(24 − r 2 )
=−
=
.
dx
4yr 2 − 24y
y(r 2 − 6)
b. A point on the curve where x = y satisfies 0 = 4x4 − 60x2 = 4x2 (x2 − 15). Since the origin
√
√
is an isolated point, there is no tangent to the curve at the origin. At the points (± 15, ± 15),
2
r = 30 and
dy 24 − 30
=
= − 14 ,
dx
30 − 6
√
√
√
so the tangent lines in are defined, respectively, by y ∓ 15 = − 41 (x ∓ 15), or x + 4y = ±5 15.
dy
c. The tangent line to the curve is horizontal where dx = 0, which is where
x(24 − r 2 ) = 0,
and
y(r 2 − 6) , 0.
If x = 0 on the curve then y 4 − 12y 2 = 0, or y 2 (y 2 − 12) = 0, and since the origin is isolated the only
√
possibility is where y 2 = 12. This gives two points, (0, ±2 3), where the tangent line is horizontal.
2
2
2
If r = 24 on the curve then 24 · 12 = 36x , or x = 8, and so y 2 = 16. This gives four more points,
√
√
√
√
(2 2, 4), (−2 2, 4), (−2 2, −4) and (2 2, −4), where the tangent line is horizontal. There are no
other points on the curve at which the tangent line is horizontal.
Review () — Solutions to ‘Another sample test’
dx = 0, which is where
d. The tangent line to the curve is vertical where dy
y(r 2 − 6) = 0,
and
Winter 
Calculus I (-nya-/)
x(24 − r 2 ) , 0.
Solution to question . — If
y = a + bx + cx2 + dx3 + x4 ,
then
dy
= b + 2cx + 3dx2 + 4x3 .
dx
There is no point on the curve where r 2 = 6 since every point on the curve besides the origin
satisfies 12 6 r 2 6 48. If y = 0 on the curve, then x4 − 12x2 = 36x2 , or x2 (x2 − 48) = 0, and since
the origin is an isolated point on the curve, the tangent to the curve is vertical where x2 = 48.
√
Hence there are two points, (±4 3, 0), at which the tangent line to the given curve is vertical.
The curve is tangent to y = −3x − 2 where x = −1 if, and only if, y = 1 and dx = −3 where x = −1,
so
a−b+c−d +1 = 1
and
b − 2c + 3d − 4 = −3.
Solution to question . — If y = tan2 (γx) then 1 + y = sec2 (γx). Differentiating twice with
respect to x gives
dy
= 2γ tan(γx) sec2 (γx),
dx
and
n
o
n
o
d2 y
= 2γ 2 sec4 (γx) + 2 tan2 (γx) sec2 (γx) = 2γ 2 (1 + y)2 + 2y(1 + y)
dx2
= 2γ 2 (1 + y)(1 + 3y).
Adding the first and third equations gives 2a + 2c + 2 = 0, or a + c = −1. Subtracting the fourth
second equation from the fourth equation gives 4c + 8 = −4, or c = −3, and therefore a = 2.
Replacing a and c by their values in the first and second equations gives −b −d = 1 and b +3d = −5.
Adding these last two equations gives 2d = −4, so d = −2 and b = 1. Therefore, the curve in
question is defined by
y = 2 + x − 3x2 − 2x3 + x4 .
Therefore,
d2 y
= 8(1 + αy)(1 + βy)
dx2
if, and only if, α = 1 and β = 3, or vice versa, and γ = ±2.
Solution to question . — a. If f (x) = x(log x)2 then
f 0 (x) = (log x)2 + 2 log x = (log x)(log x + 2),
which is zero if, and only if, either log x = 0, i.e., x = 1, or else log x = −2, i.e., x = e−2 . Therefore,
the tangent line to the graph of f is horizontal if x is 1 or e−2 .
b. If f (x) = (x − 2)2 (x2 + 2x − 12)3 then
f 0 (x) = 2(x − 2)(x2 + 2x − 12)3 + (x − 2)2 3(x2 + 2x − 12)2 (2x + 2)
= 2(x − 2)(x2 + 2x − 12) x2 + 2x − 12 + 3(x − 2)(x + 1)
= 2(x − 2)(x2 + 2x − 12)(4x2 − x − 18)
= 2(x − 2)(x2 + 2x − 12)(4x − 9)(x + 2).
The tangent line to the graph of f is horizontal where f 0 (x) = 0. Since x2 + 2x − 12 = (x + 1)2 − 13,
√
it follows that the graph of f has horizontal tangent lines where x = ±2, 49 or −1 ± 13.
Solution to question . — a. If the position of the particle is given by x = (2t − 1)3 (t − 2)6 , for
t > 0, then its velocity is
v=
dx
= 6(2t − 1)2 (t − 2)6 + 6(2t − 1)3 (t − 2)5 = 18(2t − 1)2 (t − 2)5 (t − 1).
dt
The particle is at rest when its velocity is zero; i.e., when t is 21 , 1, 2.
b. The particle is moving to the left when its velocity is negative and perhaps zero at isolated
instants. Since (2t −1)2 is never negative, and (t −2)5 and t −1 have opposite signs only if 1 < t < 2,
it follows that the particle is moving to the left on the time interval ( 1, 2 ).
c. The particle moves to the right if 0 < t < 1, to the left if 1 < t < 2 and again to the right if
2 < t < 3, so the distance travelled by the particle during the first three seconds is equal to
n
o n
o n
o n
o n
o n
o
s(1) − s(0) + s(1) − s(2) + s(3) − s(2) = 1 − (−64) + 1 − 0 + 125 − 0 = 191 metres.
dy
dy
The curve is tangent to y = 6 − 7x where x = 1 if, and only if, y = −1 and dx = −7 where x = 1, so
a + b + c + d + 1 = −1
and
b + 2c + 3d + 4 = −7.
Solution to question . — Where u and v are differentiable functions of x, so is y = u/v if v , 0,
and
dy
du 1 u dv
·
=
· −
.
dx
dx v v 2 dx
Proof. — The proof is a direct calculation apart from the use of the continuity of differentiable
functions, which insures that v 0 → v as x0 → x, and hence also that v 0 , 0 if x0 is sufficiently close,
but not equal, to x.
0
dy
y0 − y
u
u u
1
u
= lim 0
= lim
−
+
−
·
dx x0 →x x − x x0 →x v 0 v 0 v 0 v x0 − x
0
u v0 − v
u −u 1
· 0 − 0 · 0
= lim
0
0
v v x −x
x →x x − x v
du 1 u dv
=
· −
·
.
dv v v 2 dx