Solns to 123 Problem Sheet Hillary 21
24 January 2001
1. f (x) = log (1 + x2 + 3x3 ). Using the chain rule work out f 0 (x). Using the quotient rule
work out f 00 (1)
Answer:
d
(1 + x2 + 3x3 ) = 2x + 9x2
dx
and
d
1
log u =
du
u
therefore
f 0 (x) =
Next
2x + 9x2
.
1 + x2 + 3x3
d
(2x + 9x2 ) = 2 + 18x
dx
so the quotient rule gives
f 00 (x) =
and hence
(1 + x2 + 3x3 )(2 + 18x) − (2x + 9x2 )2
(1 + x2 + 3x3 )2
(1 + 1 + 3)(2 + 18) − (2 + 9)2
21
f (1) =
=
−
(1 + 1 + 3)2
25
00
2. Differenciate ln tan x with respect to x.
Answer: Once again use the chain rule and the fact that
d
1
tan x =
dx
cos2 x
to give
d
1
1
ln tan x =
=
2
dx
tan xcos x
sin xcos x
I asked this before in the Christmas exam.
1
Conor Houghton, email houghton@maths.tcd.ie
1
2
3. Use the chain rule and product rule to differenciate y = xe−x and find the extrema.
Answer:
d −x2
2
e
= −2xe−x
dx
and so
d −x2
2
2
xe
= −2x2 e−x + e−x .
dx
√
This means the differencial has zeros when x = ±1/ 2 and these are the extrema. I have
discussed this example before as well. The graph is below.
0.4
0.2
-3
-2
-1
1
x
2
3
-0.2
-0.4
4. Find the extrema of y = 2x3 + 3x2 − 12x + 9 and say weather they are maxima or
minima.
Answer: So
dy
= 6x2 + 6x − 12 = 6(x − 1)(x + 2)
dx
and hence the extrema are at x = 1 and x = −2.
d2 y
= 12x + 6
dx2
so
d2 y = 18 > 0
dx2 x=1
2
and that point is a minimum.
d2 y = −18 < 0
dx2 x=−2
and that point is a maximum. The graph is below.
50
40
30
20
10
-3
-2
-1
0
1
x
2
3
5. Using l’Hôpidal’s rule work out
e2x − e−2x
x→0
x
lim
Answer: Well the numerator and denominator are both zero for x = 0. This means we
apply l’Hopidal’s rule:
d 2x
(e − e−2x ) = 2(e2x + e−2x )
dx
and so
e2x − e−2x
= 2 lim (e2x + e−2x ) = 4
lim
x→0
x→0
x
6. Calculate the Taylor expansion of cot x around x = π/2 up to O(h3 ).
Answer: Again this is just an epic2 in differenciation:
d
1
cot x = − 2
dx
sin x
2
I won’t ask any Taylor series question this long in the exam
3
and
d 1
d2
2 cos x
cot
x
=
−
=
2
dx2
dx sin x
sin3 x
and finally
d3
sin4 x + 3cos2 xsin2
d 2 cos x
=
−2
.
cot
x
=
dx3
dx sin3 x
sin4 x
Now sin π/2 = 1 and cos π/2 = 0 means that
1
cot (π/2 + h) = −h − h3 + O(h4 )
3
4