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121
а)
x
¢+¢=¢
×
(
)
¢
¢
x
cos
¢
(
)
¢
(
)
¢×=
¢
=
(
)
¢
(
)
(
)
¢
()()(
)
¢
3
34
xxy -+= ;
Solution:
1
3
2
-+= xxxy
1
¢
b)
2
334
2
x
exy ×= sin ;
Solution:
Use the formula vuvuvu
2
2
2
-
;
234
1
-
2
( )
)( .
1
-
3
-
23
1222
2
-
2
23
124
xxxxxxxy ++=++=--×+×=
2
4
3
3
x
x
¢
( )
=
2
c)
x
y
= ;
Solution::
u
ö
Use the formula
¢
y
=
г)
æ
ç
è
-
( )
cos
2
3sin
+= xy ;
v
22
2
x
÷
ø
( )
Solution:
Use the formula uuu
uy sin
, where
u
2 x
exy +=
¢
×
10
;
д)
¢
=
( )
Solution:
Use the formula
10
uy = , where
¢
е)
+×=
x
sin2
exy
×= ;
×+×
=
xxxx
coscos +
.
2
v
¢
=
¢
¢
vuvu
-
cos
cos)(sin .
2
;
3
+= xu
2
nn
-1
¢
uunu
××=
.
x
2
;
exu +=
9
229
exexexuy +×+=
21010
2
x
23cos2cossin
×+=×=
xxx
.
sincossinsin .
2
xxxx
sincos2
xxxuuuy
.
xxxx
exexexexy ×+×=
.
122
Solution:
()(
)
¢+=¢+¢
dx
(
)
=
x
x
dx
-
cos
cos
cos
x
dx
(
)
×¢=
x
x
xxxxx
sin
sin
¢
=
sin2sin
+
Example .
Find the derivative
xx
xxxx
sin2sinsin2sin2
( )
sin2
.cos2
xexxe
2
yd
:
2
=
xexxeexexy
а)
Solution:
dy
¢
y -=
2
yd
=
б)
Solution:
xey = .
¢
¢¢
Example.
Find the differential of a function y, if
Solution:
We will use a property of logarithms to simplify formulas:
Use the formula dxydy
1
¢
y
=
æ
ctgxdy
ç
è
Find the derivative of the functions by yourself:
2
= y
y ;
-
x
xy cosln
1
¢¢
( )
-=
x
xx
xeey +=
( )
sin
-=
x
+-
xx
2
1
)(
tgxy
,
ö
÷
ø
37
;
( )
×==
cos
¢
xxy lnsinln -=
.
¢
.1dx
æ
¢
ç
è
¢
x
=
1
-=
.
22
xxxxx
.2
xeexeeey +=++=
.
cos1
1
ö
?
=
÷
2
ø
ctgx
tgx
;
xysin
ö
æ
=
ln
ç
è
11
;
-=-=-
.
÷
x
ø
x
sin
123
3
3
()2153614
=-+-
()63036
=-+
1
5
4
xy
2
xy 5
ctg=
5
ln--=
x
x
y
y =
y
y lnarccos=
-
arcsin
=
-=
1
2
x
5
2
arctg
2
=
2
tg
=
xey
1
3160
-
2
)cos( x
x
)(ln 11
-+=
ey
3
x
1
xey2
1
2
x
Topic 11. Using derivatives in curve tracing
Introduction. This workshop is devoted to consideration of finding
the asymptotes and creating a graph of functions
On the interval [a,b] the function y = f(x) can reach the lowest or the
highest values at critical points or at the ends of the segment [a,b].
Example. Find the extremum of a function
fxxxx
points of the function are x1 = 2 и x2 = 3. The extremun can be only at
these points. As you move through the point x1 = 2 the derivative changes its sign from plus to minus, at this point, the function has a maximum.
When passing through the point x2 = 3-order derivative changes sign
22
Solution: Because
.
¢
fxxx
2
= 6(x -2)(x – 3), the critical
124
from minus to plus, so at the point x2 = 3 the function reaches a mini-
1
-
x
mum.
By calculating the function values at the points x1 = 2 and x2 = 3, we
find the extrema of this function: the maximum is f(2) = 14 and the minimum is f(3) = 13.
3
Example. Investigate the function
x
=
y and draw its graph.
2
Find the domain of the function. It is obvious that this area is
(-¥; -1) È (-1; 1) È (1; ¥).
In turn, we can see that the lines x = 1, x = -1 are the vertical asymptotes of the curve.
The domain of this function is the interval(-¥; ¥), exept -1 and 1 .
The actual range of this function is the interval (-¥; ¥).
The points of discontinuity of a function are the points х = 1, х = -1.
Find the critical points.
First of all let's find the derivative of a function
322
2)1(3
×--
¢
=
y
-
x
xxxx
22
)1(
=
-
x
424
233
--
xxx
=
22
)1(
24
3
-
xx
22
)1(
-
x
The critical points are: x = 0; x = -3; x =3; x = -1; x = 1.
Find the derivative of a function of the second order
224223
-----
)1(4)3()1)(64(
¢¢
=
y
=
x
=
x
35
642
-+
xxx
=
-
x
=
42
)1(
24
-+
xxx
42
)1(
-
x
42
-
)1(
x
42
-
)1(
42
-
)1(
)32(2
=
xxxxxxx
xxx
)1(
-
x
=
324243
22
42
---+--
)44)(3()12)(64(
xxxxxxxx
=
355735357
-++--+-+-
1212446126484
xxxxxxxxxx
=
2
)1)(3(2
-+
=
x
)3(2
+
xx
32
)1(
-
Investigate the convexity and concavity of the curve at intervals.
.
-¥ < x < -3, y¢¢ < 0, the curve is convex (concave down)
3
-
< x < -1, y¢¢ < 0, the curve is convex
-1 < x < 0, y¢¢ > 0, the curve is concave up
125
0 < x < 1, y¢¢ < 0, the curve is convex
x
1 < x <
3
3
, y¢¢ > 0, the curve is concave up
< x < ¥, y¢¢ > 0, the curve is concave up
Find the intervals of increasing and decreasing functions. For this we
determine the signs of the derivative of the function on the intervals.
-¥ < x < -3, y¢ > 0, function increases
3
-
< x < -1, y¢ < 0, function decreases
-1 < x < 0, y¢ < 0 function decreases
0 < x < 1, y¢ < 0, function decreases
1 < x <
3
3
, y¢ < 0, function decreases
< x < ¥, y¢¢ > 0, function increases
Evidently, point х = -3 is a maximum point, and х = 3 is a min-
imum point. The function values at these points are equal to -33/2 and
33/2 respectively.
We have already talked about vertical asymptotes. Now let’s find in-
clined asymptotes.
2
x
lim
=
k
3
æ
x
ç
lim
=
b
ç
è
-
x
2
1
-
x
æ
ö
ç
÷
lim
=
ç
÷
è
ø
=
2
1
-
x
33
1
-
x
+-
lim
xx
xxx
1
;1
¥®¥®
=
1
1
-
2
1
ö
÷
=
÷
ø
lim
x
=
22
1
-
x
lim
xxxx
x
0
¥®¥®¥®¥®
1
=
1
-
2
x
Thus we get the asymptote equation – y = x.
Let’s draw the function graph:
126
3
4
1
-
x
(
)
=
¢
Î
=
Ï-=
()()(
)
2
1
-2 -1 1 2
-1
-2
-3
-4
Fig.3
3
Fig.3 – graphic representation of the function
x
=
y
2
Example .
Find the largest and smallest value of the function
3 xxy -= in the
3
segment [0; 3].
Solution: The function reaches its highest and lowest values in stationary points of a given segment, or at the ends of that segment. Let’s
find the stationary points (the points, where the derivative is equal to 0 or
doesn’t exist):
¢
0
y
if
x
22
1333 xxy -=-=
]3;0[1
and
x
]3;0[1
Let’s find the function’s values at these points and at the ends of the
given segment
183;00;21 -=== yyy
Now pick the smallest and the largest value of the ones we got.
127
So, the largest value on the given segment is 2 and is reached at х = 1,
(
)
==-
()(
)
(
)
y
наиб
y
наим
2)1( =
, and the lowest value is equal to -18 when х = 3,
.18)3( -=
Example .
3
2
+
x
=
Investigate the function
y
( )
and draw the graph.
2
14
-
x
Solution:
The general algorithm to investigating functions is:
· Find the function’s domain.
· Investigate the behavior of a function at the ends of the domain.
Find the break points of a function as well as the one-sided limits at
these points. Find the vertical asymptotes.
· Find out whether the function is even, odd or periodic.
· Find the crossings with the axes of the graph, and the intervals of
sign consistency.
· Find the inclined asymptotes on the function’s graph.
· Find extremum points and intervals where the function increases
and decreases
· Find inflection points of the function’s graph and its intervals of
convexity and concavity.
· Draw a schematic graph of the function, using all this information.
)1(,01
1. The function is not defined, if
And the domain is:
¥È¥-Î ;11;x
xx
2. Because х = 1 is the function’s break point, we need to investigate
the function’s behavior at this point from the left and the right
3
+
2
lim
x
lim
x
-®
01
+®
01
x
( )
-
x
( )
+
x
( )
x
+¥=
,
2
14
3
2
+¥=
2
-
14
х = 1 is the point of discontinuity of the second kind, because the limits are equal to ∞.
Therefore, the line х = 1 is a vertical asymptote.
Let’s check whether the function is even, odd or neither. Remember,
that function у = f(x) is called even (odd) if two conditions are met:
The domain is symmetrical about the point of origin
128
)).()(()()( xfxfxfxf
-=-=-
(
)
(
)
-==
xfy
+
=
x
[
]
(
)
4
If у = f(x) is even, the graph is symmetrical about the y-axis, and if
odd – about the point of origin.
( )
xf
=-
2
+-
x
( )
14
--
x
3
-=
2
2
-
x
( )
14
+
x
3
2
This function is neither even nor odd, i.e. it has a general form.
The function is not periodic. Let’s find the crossover points of the
graph with the axes
;2i0:OXwith
;2i0:OYwith
==
yfx
Let’s find the function’s sign continuity
3
)2(
+
x
0
Þ<
y
0
Þ>
y
2
)1(4
-
x
3
)2(
+
x
2
)1(4
-
x
xx
5. Let’s find the inclined asymptotes
xf
±¥®
+¥®
+¥®
xy
x
é
( )
x
ê
( )
ë
+=
)(
2
+
( )
14
-
xx
2
+
14
-
x
kxxfb
¥
¥
x
ù
ú
û
ù
ú
û
-
)(lim
1
;
=
4
[ ]
2
)1(
1
lim
=¥-¥=
xx
4
2323
28126
-+-+++
xxxxxx
x-=±¥®
3
é
=
2
ê
ë
3
1
-
2
4
x
asymptote inclined theis 2
k
lim
=
x
lim
=
k
x
lim
=
b
1
lim
=
x
4
1
xx
+¥®+¥®
é
=
ê
ë
);2;(020
--¥ÎÞ<+Þ<
);1()1;2(020
¥+È-ÎÞ>+Þ>
,bkxy
where
23
)1()2(
--+
xxx
=
2
)1(
-
x
¥
ù
2
=
ú
¥
û
For x → -∞ k and b are calculated similarly
6. Let’s find the extremum points and intervals of increasing and de-
creasing of the function.
Increasing and decreasing of the function у = f(x) is characterized by
the sign of its derivative y': if at any interval y' > 0, then the function is
increasing at this interval, and if y' > 0 – it is decreasing.
Function у = f(x) can have extremum points inside its domain and
when the derivative is 0 or cannot exist. If y' changes its sign from “+” to
129
“-” by passing this point, then it is a maximum, if y' changes its sign
()()()(
)
()(
)
х (-∞; -2) -2 (-
2; 1) 1 (1; 7)
7
(7; +∞)
y
'
doesn’t
y
0
doesn’t
d
e-
144
from “-” to “+” by passing this point, then it is a minimum. If y' doesn’t
change its sign by passing this point, it is not an extremum.
Let’s find all points inside the domain of y = f(x), where the derivative (y') turns to 0 or doesn’t exist.
322
212123
+---+
¢
=
y
¢
¢
( )
14
-
x
=-==
xxfy
21
xy
=
xxxx
4
;7,2i0
1if exist,t doesn'
=
2
xx
( )
14
-
x
72
-+
.
3
Let’s create a table
+ 0 +
exist
exist
- 0 +
≈ 5
increase increase
crease
min increase
The function increase at intervals (-∞; -2), (-2; 1), (7; +∞) and de-
729
)7(
scends at (1; 7). Point х = 7 is the minimum
min
== yy
.
7. Let’s find inflection points of the function’s graph and its intervals
of convexity and concavity.
A reminder: the graph of function y = f(x) is convex at an interval
);( ba
, if at any point of this interval the graph lies under any of its tar-
get lines. The graph of function y = f(x) is concave at an interval (а; b), if
at any point of this interval the graph lies above any of its target lines.
130
()()(
)
[
]
()()()(
)
(
)
х
-2
1
y'' - 0 + Doesn’t exist
+ у
Ç
0
È
Doesn’t exist
È
(
)
x x
y y
0 0
convex function concave function
Fig.4
Fig.4 – graphic representation of convex and concave function.
Points, at which the functions changes concavity to convexity and
vice versa, are called flex points (inflection points).
A flex is possible at points, where y'' is equal to 0 or can’t exist. If
y'' < 0 on an interval (а; b), then the graph is convex (Ç) at this interval,
if y'' > 0, then at interval (а; b) the graph is concave (È).
Let’s find flex points y = f(x):
2232
721312722
-+---++-+
¢¢
=
y
¢¢
¢¢
2if0
-==
xy
=
( )
1x if exist,t doesn' y
14
-
x
xxxxxxx
=
+
x
( )
14
-
x
254
+
x
=
)1(2
-
x
Let’s make a table
(-∞; -2)
(-2; 1)
(1; +∞)
)2(27
446
Point (-2; 0) is a flex point.
Extra points:
01,03
()
»
( )
-»-
8,73
3,06
-»-
y
y
y
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