Download Solutions to Day 6 Tutorial

The University of Sydney
School of Mathematics and Statistics
Solutions to Day 6 Tutorial
MATH1003: Integral Calculus and Modelling
Summer School 2016
Web Page: http://www.maths.usyd.edu.au/u/UG/SS/SS1003/
Lecturer: Dr Bob Crossman
This tutorial covers material from the lectures in Day 5.
1. Simplify the expression ex+ln x .
Solution: ex+ln x = ex eln x = xex .
2. Find dy/dx for each of the following:
1+x
(a) y = ln
1−x
2
Solution:
1 − x2
(b) y = log2 x
1
Solution:
x ln 2
2
(c) y = 3 log2 (x )
Solution: 3 log2 (x2 ) = 6 log2 (x) =
6 ln x
dy
6
, so
=
.
ln 2
dx
x ln 2
(d) y = xx
d x
d x ln x dy
=
(x ) =
= ex ln x x ×
e
Solution:
dx
dx
dx
√
x2 3 7x − 4
(e) y =
(1 + x2 )4
1
x
+ ln x = xx (1 + ln x).
Solution: ln y = 2 ln x + 31 ln(7x − 4) − 4 ln(1 + x2 ).
Differentiating both sides with respect to x, we have
1 dy
2 1
7
2x
= + ×
−4×
.
y dx
x 3 7x − 4
1 + x2
√
dy
x2 3 7x − 4 2
7
8x
Therefore,
.
=
+
−
dx
(1 + x2 )4
x 3(7x − 4) 1 + x2
c 2015 The University of Sydney
Copyright 1
3.
(a) Use l’Hôpital’s rule to find lim ln (1 + x)1/x .
x→0
1
ln(1 + x)
= lim 1+x = 1, where l’Hôpital’s
Solution: lim ln (1 + x)1/x = lim
x→0 1
x→0
x→0
x
rule has been used in the second step.
(b) Hence find lim (1 + x)1/x .
x→0
[Note: lim ln (1 + x)1/x = ln lim (1 + x)1/x . Can you say why?]
x→0
x→0
Solution: Since ln x is continuous,
lim ln (1 + x)1/x = ln lim (1 + x)1/x = 1.
x→0
x→0
Therefore, lim (1 + x)1/x = e.
x→0
4.
√
(a) For which positive real numbers x is it true that x > x/2?
√
Solution: If x > x/2 then 4x > x2 or (4 − x)x > 0 which requires
0 < x < 4.
(b) Without using your calculator,
that π ≈ 3, determine which
√
√ and assuming
of the following is bigger: ( π)π or π π .
[Hint: The exponential function is always increasing, so if a > b, then ea > eb .]
√
√
√
√
Solution: ( π)π = exp(π ln√ π) = exp( π2 ln π), and π π = exp( π ln π).
Now, 0 < π < 4, and
so π2 < π (by part (a)).
√
√
Therefore π2 ln π < √ π ln π (ln π > 0), and exp π2 ln π < exp( π ln π).
√
That is, ( π)π < π π .
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