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Solutions Sheet 3, Di erentiation Questions 29-44
Solutions to Additional Questions
2
29) Set f (x) = sin (3 +x) , then
2
f (x) f (0) sin (3 +x) sin 9
= :
x 0 x
So
2sin (3 +x) sin 9
0lim =f (0):
x!0 x
2
The function f (x) is the composite of siny and y = (3 +x) ; both of
which are di erentiable. By the composite rule for di erentiation we have
df (x) d sinydy
= = cosy 2 (3 +x)
dx dy dx
2= 2 (3 +x) cos (3 +x) .
Hence
2 sin (3 +x) sin 9 2lim = lim 2 (3 +x) cos (3 +x) = 6 cos 9:
x!0 x!0x
329) Rewrite as f (x) =jxj as
8
3< x x> 0
f (x) = 0 x = 0
: 3x x< 0:
The only di culty is checking di erentiability at x = 0. I leave it to the
student to check both one-sided limits of
f (x) f (0)
x 0
as x! 0+ and x! 0 and thus show
8
23x x> 0<
0f (x) = 0 x = 0:
: 23x x< 0
1Similarly, 8
6x x> 0<
00f (x) = 0 x = 0;
:
6x x< 0
00 00or f (x) = 6jxj. From the notes we know thatjxj, and thus f (x), is not
di erentiable at x = 0.
31) i) The de nition of di erentiable involves a limit. We will consider the
two one-sided limits.
Consider rst x> 0, when
2 2g (x) g (0) x sin (=x ) 0
= ;
x 0 x 0
Thus
g (x) g (0)
lim = lim x sin :
2x!0+ x!0+x 0 x
By the Sandwich Rule this limit exists and equals 0.
Next, for x< 0 we have
g (x) g (0) 0 0
= = 0:
x 0 x 0
Thus
g (x) g (0)
lim = lim 0 = 0:
x!0 x!0x 0
Since the two-sided limits exist and are equal we deduce that
g (x) g (0)0g (0) = lim
x!0 x 0
exists and equals 0:
ii).For non-zero x we simply use the rules of di erentiation to get
2 0g (x) = 2x sin cos :
2 2x x x
iii)
p p1 20g p =p sin (2n ) 2 2n cos (2n ) = 2 n; (1)
2n 2n
2since sin (2n ) = 0 and cos (2n ) = 1 for all n2N.
0iv) Assume that lim g (x) does exist, with value ‘ say.x!0
Choose " = 1 in the de nition of the limit to nd > 0 such that
00<jxj< impliesjg (x) ‘j< 1, i.e.
0 0jg (x)j = jg (x) ‘ +‘j
0 jg (x) ‘j +j‘j
by triangle inequality
< 1 +j‘j: (2)
p
But if we choose n2N su ciently large and set x = 1= 2n we can havenp
0both 0 <jxj < and, by (1);jg (x )j = 2n > 1 +j‘j. This contradictsn n
(2) and thus our assumption, that the limit exists, is false.
32) Let
2xxf (x) =e 1 x ;
2
0 xnoting that f (0) = 0. Thenf (x) =e 1 x which, by the assumption in
the question, is > 0 for all x = 0. Assume x < 0, then by the Mean Value
theorem there exists x<c< 0 for which
0f (x) f (0) =f (c) (x 0):
0This is < 0 since x < 0 and f (c) > 0. Thus f (x) < f (0) = 0 which
rearranges to
2xxe 1 +x + :
2
33) Let
x
f (x) =p arcsinx
21 x
for 0<x< 1; noting that f (0) = 0. Then
p
2x2 p1 x 2 1 x0 2 1 xf (x) = p =p > 0
2 2 21 x 1 x 1 x
for our x. The Mean Value theorem implies that there exists x < c < 0 for
which
0f (x) f (0) =f (c) (x 0)> 0:
3
6This rearranges to the required result.
x 234) i) Letf (x) = 2 x and look for some sign changes. Randomly choosing
values forx leads tof ( 1) = 3=4,f (0) = 1, f (1) = 1, f (2) = 0 (giving a
solution!) f (3) = 1 and f (4) = 0 giving another solution. So other than
the solutions at x = 2 and 4 the Intermediate Value Theorem gives a third
solution between 1 and 0.
ii) To show that it has exactly three solutions we assume, for a contra-
diction, that it has more, i.e. at least four. Then by the result in Question
3(3) (3) x8 there exists a c2R for which f (c) = 0. In this case f (x) = (ln 2) 2
which is never zero. This contradiction means the function has at most 3
solutions. Since we know it has at least 3, we conclude it has exactly 3
35) Apply the Cauchy Mean Value Theorem to f and g (x) = lnx. This is
allowable since x2 [a;b] and it is being assumed that a > 0. Then there
exists c2 [a;b] for which
0f (b) f (a) f (c)
= :
lnb lna 1=c
this rearranges to the stated result.
36) From an earlier question we have
d 1 1
(arcsinx + arccosx) =p p = 0
2 2dx 1 x 1 x
for x 2 ( 1; 1). By the Increasing-Decreasing Theorem this means that
arcsinx + arccosx is constant on ( 1; 1). To nd its value take x = 0; when 0 + 0 ==2. Hence
arcsinx + arccosx =
2
for x2 ( 1; 1).
Similarly
1d 1 1 1 12uarctanu + arctan = + = = 0:12 2 2du u 1 +u 1 + 1 +u 1 +u2u
41So arctanu + arctan is constant. Take u = 1 to see that
u
1
arctanu + arctan = 2 arctan 1 = 2 = :
u 4 2
for all u> 0.
37) i) For x = 0 but near 0,
x cosx sinx x cosx x +x sinx
=
3 3x x
x cosx x x sinx
= +
3 3x x
cosx 1 x sinx
= +
2 3x x
1 1 1
! + = ;
2 6 3
as x! 0, using assumptions in the question.
ii) For x = 0 but near 0,
tanx x sinx x cosx 1 x cosx sinx
= =
3 3 3x x cosx cosx x
1 1 1
! = ;
1 3 3
as x! 0, using the result from part i.
iii)
3 3tanx x x tanx x x tanx x
3= = cos x
3 3 3 3tan x tan x x sinx x
1 1
! 1 1 =
3 3
as x! 0, using part i) along with results from lectures.
iv)
sin 3x 3x sin 3x 3x
= 27 :
33x (3x)
3If we write f (x) = (sinx x)=x when x = 0 and f (0) = 1=6 then
sin 3x 3x
=f (3x)3
(3x)
5
666is the composition of two continuous functions. Thus
1
limf (3x) =f lim 3x =f (0) = :
x!0 x!0 6
Hence
sin 3x 3x sin 3x 3x 27
lim = 27 lim = = 4:5:
33x!0 x!0x 6(3x)
v)
sin 3x 3 sinx sin 3x 3x + 3x 3 sinx
=
3 3x x
sin 3x 3x sinx x
= 27 3
3 3x(3x)
1 1
! 27 3 = 4;
6 6
by part iii).
38) 8
x cosx sinx< if x = 0
0 2xf (x) =
:
0 if x = 0;
(2)For f (0) consider
0 0f (x) f (0) x cosx sinx
= :
3x 0 x
This has been seen in the previous question, where it was shown to have limit
1=3.
39) i) From an earlier question,
d 1 1 sinhy
arcsin = r
2 dy coshy 2 cosh y
11
coshy
coshy sinhy 1
= = :2sinhy cosh y coshy
6
6This holds when not only is sinhy = 0 but we must restrict toy for whichp
2sinhy is positive since we take the positive square root in cosh y 1 =
sinhy. Thus result holds for y> 0.
ii) From an earlier question, we have for all y2R.
d 1 coshy
(arctan (sinhy)) = coshy =
2 2dy cosh y1 + (sinhy)
1
= :
coshy
iii) From an earlier question,
d 1
parccosy =
2dy 1 y
for anyy2 ( 1; 1). We could replacey by 1= coshy as done in part (i), and
I leave that to the interested Student.
Alternatively, replacey by tanhy since we know that tanhy2 ( 1; 1) for
all y. Then
d 1 1 1
arccos (tanhy) = q = coshy
2 2dy 2 cosh y cosh y
1 (tanhy)
1
= :
coshy
Valid for all y2R.
40) i) Let f (x) = ln (1 +x). Then
1(1) (1)f (x) = (1 +x) ; so f (0) = 1;
2(2) (2)f (x) = (1 +x) ; so f (0) = 1;
(3) 3 (3)f (x) = 2 (1 +x) ; so f (0) = 2;
4(4) (4)f (x) = 3! (1 +x) ; so f (0) = 3!;
5(5) (5)f (x) = 4! (1 +x) ; so f (0) = 4!:
7
6Thus the rst 6 approximations to ln (1 + x), i.e. T (ln (1 +x)) forn;0
0n 5 are
T (ln (1 +x)) = 0;0;0
T (ln (1 +x)) = x;1;0
2x
T (ln (1 +x)) = x ;2;0
2
2 3x x
T (ln (1 +x)) = x + ;3;0
2 3
2 3 4x x x
T (ln (1 +x)) = x + ;4;0
2 3 4
2 3 4 5x x x x
T (ln (1 +x)) = x + + :5;0
2 3 4 5
Choosing x = 0:5 we get a sequence of approximations to ln 2 of
1; 0:5; 0:83; 0:583; 0:783; 0:616; ::::
This is very slow convergence.
n(n)ii) From above we see that for eachn 1,f (x) = (n 1)! (1 +x) ; so
(n)f (0) = (n 1)!: Thus the Taylor series for ln (1 +x) is
2 3 4 5 6x x x x x
x + + +:::
2 3 4 5 6
which converges for 1<x 1.
Replace x by x in the Taylor series for ln (1 +x) to get
2 3 4 5x x x x
ln (1 x) = x ::: ,
2 3 4 5
valid for 1x< 1. Note that
1 +x
ln = ln (1 +x) ln (1 x):
1 x
We would like to obtain the Taylor series forg (x) = ln ((1 +x)= (1 x))
by subtracting that for ln (1 x) from the one for ln (1 +x). But you need
to justify the subtraction of in nite series. To calculate the Taylor series for
8(n)g we need to calculate g for all n 1. But g (x) = ln (1 +x) ln (1 x)
(n)=f (x) h (x), say, so g can be found as the di erence of the derivatives
thof f and h or, in other words, the n -term for ln ((1 +x)= (1 x)) is the
thdi erence of the n -terms for f andh. So we are allowed to subtract term-
by-term to get
3 51 +x 2x 2x
ln = 2x + + +:::
1 x 3 5
for 1<x< 1.
iii) Put x = 1=2 in ln (1 x) to get approximations to ln 2 of
0:5; 0:625; 0:6; 0:68229:::; 0:68854:::; 0:6911458::::; :::
iv) Put x = 1=3 in ln ((1 +x)= (1 x)) to get approximations to ln 2 of
0:6; 0:69135:::; 0:69300:::; 0:69313:::; 0:693146:::; 0:693147::::;: ::
Note ln 2 = 0:69315::: and in each case above we are getting sequences
that converge quicker than in the preceding case.
41) There is no need to calculate the Taylor polynomial for sinx, just La-
(6)grange’s form of the error. So with f (x) = sinx we have f (x) = sinx
and
sinc 6R f (x) = x5;0
6!
for some c between 0 and x. Butjsincj 1 and so, withjxj 0:25 we nd
6(0:25)
7jR f (x)j = 3:390844::: 10 : (3)5;0
6!
To nd the actual error we do need the Taylor polynomial
3 5x x
T f (x) =x + :5;0
6 120
The value at 12 or =15; is
3 5 1 1
T f = +5;0
15 15 6 15 120 15
0:2079116943::: .
9The di erence between the value of the Taylor polynomial and the true
9value of sin (=15) is 3:505219:::10 ; smaller, which was to be expected,
than the bound in (3).
1=342) If f (x) =x then
0 2=3f (x) = x =3;
00 5=3f (x) = 2x =9;
000 8=3f (x) = 10x =27:
0 00When a = 8, then f (8) = 2, f (8) = 1=12, and f (8) = 1=144; so
2
(x 8) (x 8)
T f (x) = 2 + :2;8
12 288
The error, in Lagrange’s form, is
000f (c) 3R f (x) = (x 8)2;8
3!
for some c between 8 and x. We are told to restrict to x2 [7; 9].
If x> 8 then R f (x)> 0 but also 8<c<x< 9 and so2;8
3
10 (x 8) 10 10
R f (x) = < = < 0:000241127:2;8 8=3 8=3 827 3!c 27 3! 8 27 3! 2
If x< 8 then R f (x)< 0 but also 7<x<c< 8 and so2;8
3
10 (x 8) 10
R f (x) = > > 0:000344263:2;8 8=3 8=327 3!c 27 3! 7
(1)43) i) By the hint given f (x) = 2g (x) in which case, from looking back
at the earlier question,
(2 (n + 1))!n(n) (n+1)f (0) = 2g (0) = 2 ( 1)
n+14 (n + 1)! (2 (n + 1) 1)
2 (n + 1) (2n + 1) (2n)!n
= 2 ( 1)
n+14 (n + 1)n! (2n + 1)
(2n)!n
= ( 1)
n4 n!
10
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