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Differentiation
3
11.1 Review
,
2
11.2 Higher Derivatives
3
11.3 Differentiatinga Product
5
11.4 Differentiating a Quotient
6
11.5 Differentiatingeaxand In x
'
7
11.6 The SecondDerivativeTest
9
FurtherExercises
11
Answersto Exercises
12
@ J.Carr/HWU/1996
,
:
Differentiation3
Review
"f
I
Page2
j
:
11.1 Review
;
.j
:.f
In the units Differentiation 1 and 2 somerulesfor differentiatingweregiven. In this
sectionwe rapidly review theserules. The derivativesof xn, sinax and cosax are
given on the formula sheet,andarelistedbelow:
f(x)
xn
sin ax
cosax
f'(x)
nxn-l
acosax
-asinax
(1.1)
In the formulae (1.1) 'a' is a constant. Recall also that, if y = f(x), then f'(x),
dy/dx and y' are all equivalent notations for the derivative.
Example 11.1 Differentiate
(i) f(x)
=
x3
- 7~2 + 3x + 2
(ii) Y
= 2sinSx - 3cos4x .
Solution
(i) f'(x)
= 3X2- 14x+ 3 usingtherulefor differentiating xn.
(ii) Using (1.1) I
~=
10cosSx+ 12sin4x
dx
-
!
f
j
...
The chainrule (seesections2 and 3 of Differentiation2) is usefulfor differentiating
compositionsof functions. If y is a functionof u , and u is a function of x, then:
!
dy
dy
du
dx
du
dx
-=-x-
Example 11.2 Find
~
if
Y = (X2 - 3x +'6)5. .
dx
Solution
y=u 5 where u=x 2 -3x+6.
Hence
-dy = -dy x -du
dx
du
dx
= SU4(2x - 3)
= S(X2- 3x + 6)4(2x - 3) .
.
11.1
Exercises
1.
Differentiate(i) X4- 7X3+ 6
2.
Find f'(x) if f(x) =
~.
X
3.
Find
~
dx
(ii) , Scos3x- sinK .
,
\
when (i) y = (x2- x + 3)7 (ii) y = (X2 + 1)-1 (iii) y = x3 + (X3 + 1)-2 .
1
j
4.
If
y = (2x + 1)-.1/2 show that y':;: _y3 .
~
j
!
4
J
..
:.
Differentiation 3
Higher Derivatives
Page3
'\,
11.2 Higher Derivatives
.-
"
.
For y = f(x), the derivative f'(x) measures
the gradientof the graphof y. The
secondderivativemeasureshow that gradientchangesandis calculatedby
differentiating f '(x) .
The following notationsareusedfor the secondderivative:
.
f "(x) (readas II f double-dashof x ")
.
d2
-{
(readas II deetwo y by dee x squared")
dx
.
Thus if
y" (readas II y double-dash") .
y = f(x) = X4 ,
y' =
~
= f '(x) = 4X3.
dx
d2
y" = -1dx2
= f"(X)
= -d (4X3) = 12x2 .
dx
Example11,3 Find y" if (i) Y = 5sin3x
(ii) y
= 10x3-
-
7X2+ 2x
Solution
(i)
y'
= 15cos3x so y" = -45sin3x
(ii) y' = 30x2-14x+2
so y" = 60,,:-14 .
Example11.4 If Y = 3cos2x showthat
d2y + 4y = 0
~
dx
Solution
.
-dy = -6sm2x
dx
d2 +4y
-{
dx
-d2y = -12cos2x
dx2
= -12cos2x
+ 4(3cos2x)
= 0 .
.
,
Differentiation3
Higher Derivatives
Page4
- ~~~---
Example 11,5
-"
..I
= Asinx + Bcosx, where A and B are constantsand f'(O) = 3,
If f(x)
f "(0) = -2, find A and B.
Solution
f'(x) = Acosx - Bsinx
f"(x) = -Asinx - Bcosx
f'(O) = 3
so Acos(O)
-
Bsin(O) = 3 .
Since cos(O)= 1 and sin(O)= 0, wehavethat A = 3 .
Using f"(O) = -2:
-Asin(O) - Bcos(O)= -2
-A
-
x0
Bx 1 =
so B = 2 .
-B
The third derivativeof y = f(x) is found by differentiating f "(x) andis denotedby
f "'(x),
dJ
-{
y'" or
. Thus if y = 31 then:
dx
x
d (x-J) = -3X-4
y' = -dx
-
;
!
y" = 12x-5
y"l =
-60X-6
':
;)
Exercises
11.2
1.
Find y" when (i) y = 4cos2x
2.
If
3.
If
y=2sin3x
f(x)
= xJ -
(ii) y = XJ_2X2+3x+4
showthat y"+9y = 0
7X2 find f"'(x)
.
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Differentiation3
Differentiatinga Product
Page5
"
11.3 Differentiating a Product
The function f(x) = x2sinx is the productof the functions g(x) = x2 and
h(x) = sinx , that is f(x) = g(x)h(x) . It is easyto find g'(x) = 2x and
h'(x) = cosx. We canfind f'(x) from the formula:
f'(x)
d
= -(g(x)h(x)
dx
= g'(x)h(x) + g(x)h'(x)
(3.1)
Formula(3.1) is calledthe product rule. For f(x) = x2sinx, by (3.1):
f'(x) = 2xsinx + X2cosx
Example11.6 Find f(x) when (i) f(x) = (x+ 1)5x3 (ii) f(x) = (X2+ 1)4cosx .
Solution
(i) f(x) = g(x)h(x) where g(x)= (x + 1)5 and h(x)= x3 .
Bytheproduct
rule:
'-
f'(x) = g'(x)h(x) + g(x)h'(x) = 5(x + 1)4x3+ (x + 1)5(3x2).
(ii) f(x)
= g(x)h(x) where g(x) = (X2+ 1)4 and h(x) = cosx .
f'(x) = g'(x)h(x) + g(x)h'(x)
= 4(X2+ 1)3(2x)cosx+ (X2+ 1)4(-sinx)
=
8x(X2+ 1)3cosx - (X2 + 1)4sinx .
Example11.7 Showthat y
= xsinx satisfies y" + Y = Acosx
,
where A is a
constantandfind A.
Solution
Usingtheproductrule: y' = sinx + xcosx.
Differentiating
againandusingtheproductruleforthesecondtermgives:
y" = cosx+ COSX
- xsinx = 2cosx- xsinx .
Then
y" + Y = 2cosx - xsinx + xsinx
= 2cosx
so A = 2.
Exercises 11.3
(i) f(x) = X3cosx (ii) f(x) = (4x + 3)(x + 2)5.
1.
Find f'(x) when
2.
Find y' when y = (X2+ 1)l/ix + 3) .
3.
Showthat y = xsin3x
satisfies
~
dx
constant,and find A.
+ 9y
= Acos3x
where A is a
~
'I
Differentiation3
Differentiatinga Quotient
',,':
--
11.4 Differentiating a Quotient
I
,
If
f(x) =
~
Page6
;)1';
-
"';t'1-",'.\
.I
i!.-
~'
then we canfind f '(x) from the formula:
h(x)
f'(x) = g'(x)h(x)-g(x)h'(x)
r
(4.1)
h2(x)
This is calledthe quotient rule.
Thus if
f(x)
=
~
then f(x) =
x
~
, where g(x) = sinx, h(x) = x3. Then:
h(x)
f'(x) = g'(x)h(x)-g(x)h'(x)= x3cosx-3x2sinx.
h2(x)
Example 11.8 Find f'(x)
!
~
Solution
I
.f(x) =
f'(x)
.
x6
f(x) = 2x+1
3x+4
when
' where g(x) = 2x + 1 , h(x) = 3x + 4. Then:
-
= g'(x)h(x)-g(x)h'(x)
h2(x)
-- 2(3x+4)-3(2x+l)
(3x+4)2
-- 6x+8-6x-3
(3x+4)2
x+1
Example11.9 Find f '(x) when f (x) = -y-
5
-- ~-:;:4)2
.
.
x +1
Solution
f(x)
=
~
, where g(x) = x + 1 , h(x) = X2+ I.
h(x)
f'(x) = g'(x)h(x)-g(x)h'(x)
h2(x)
-- (x2+1)-2x(x+l)
-- x2+1-2x2_2x
(X2 + 1)2
(X2 + 1)2
Then:
-- -x2-2x+1
(X2 + 1)2
Exercises 11.4
1.
.
Fmd
f'(x) when f(x)
2.
Find f '(x) when f(x) =
=
(X2 +1)
(x + I)
~
.
:'
.
smx
3.
Let f(x) = ~.
Express f'(x) in the form A(5x - 1)-2 where A is a
5x-1
constant,andhencefind f"(x).
1
--
..:
~
i
Differentiation 3
Differentiating eax and lnx
Page 7
'9,-
-
11.5 Differentiating
- "_c _c, ,--"
..
eax and In x
Therulesfor differentiatinglnx and ea.xare:
~ (lnx) =.!.,
dx
~
x
(eax) = aeax
(5.1)
dx
where a is a constant.
Thus if y = e3x then y' = 3e3x.
Examp!e11.10 Find f '(x) when (i) f(x) = e-2x (ii) f(x) = xe5x (iii) f(x) = exp(-x1 .
-"
~{
~;tg!,~
;~~f};,,;;,;
S I t'l
I
I
0 u on
(i) Using (5.1) with
a=-2
:
2
f(x) = g(x)h(x) with g(x) = x , h(x) = e5x so:
(ii) Using the product rule
!
gives f'(x)=-2e-x
f'(x) = g'(x)h(x)+ g(x)h'(x)
= e5x+ 5xe5x
(iii) f(x)
= e-u
where u = X2, so by the chain rule:
f '(x) =
-e-U(2x)
= -2xexp(-x1 .
Example 11.11
Find
f"(x) when
f(x) = x(lnx + 1)
Solution
Using the product rule
f'(x)
= (lnx+l)+x(~)
Thus f"(x) = -1
x
= lnx+2
.
Example11.12 Showthat y = e-3x satisfiesan equationof the form:
y" + 6y' + Ay
=
0
where A is a constant,and find the valueof A. Showalso that,with this valueof A ,
Y = xe-3x also satisfiesthe equation.
I
~
,
Differentiation
3
Differentiating
eax
and
Inx
Page
-,
Solution
= e-3x
y
y"
+
6y'
so y'
+
= -3e-3x
=
Ay
and y"
ge-3x
+
8
:
_./
= ge-3x
6(-3e-3x)
+
. Then:
Ae-3x
= e-3X(A- 9) = 0 , if A = 9 .
If Y = xe-3x,then, usingtheproductrule:
y'
= e-3x- 3xe-3x
y"
= -3e-3x- 3[e-3x- 3xe-3x]
= -6e-3x
Then
y" + 6y' + 9y =
+ 9xe-3x .
+ 9xe-3x+ 6e-3x- 18xe-3x+ 9xe-3x
-6e-3x
= O.
,
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Exercises
j',.
11.5
u
~
\ ~
Differentiate
(i)
e-5x
+
e3x
(ii)
..'
etp(
' ",f
\,"':
".
t
'
-
... -
-2X2)
--
~
1.
",c
J,
I
;.
$;:;..i""
'-
,
2.
If Y= xe2xfind y'.
~.~
!}'}"
1
I
3.
Differentiate
(i) 3lnx
~
(ii) In(3 + x)
4.
If Y = e4xsinx showthat
-d2y - 8--.,-dy + 17y = 0
dx2 dx
5.
Show
that
y
=
ex
satisfies
an
equation
y"
where
A
Show
6.
If
is
that,
f(x)
a
constant,
with
=
x
and
this
-
In(
1
+
+
find
value
of
x2)
Ay'
A,
+
the
+
Y
form:
=
0
A.
show
(1
of
y
=
xex
also
satisfies
that
x2)f'(x)
=
!
..
!
i
t
,
1
!
oi
.J
(1
-
x)2
.
the
equation.
1
.
~
"
Differentiation 3
The SecondDerivativeTest
Page9
-
'~
11.6 The SecondDerivative Test
A stationarypoint of a function f(x) is a point x = c wherethe gradientis zero,that
is f'(c) = 0 (seethe unit Differentiation2 (10.5)). In figure 1 the stationarypoint
at x = a is a local minimum,while the stationarypoint at x = b is a local maximum.
w,
t
.ty
I
Figure I
The natureof a stationarypoint at x = a canbe foundby finding the sign of f '(x)
near x = a. Thusthe stationarypoint x = a is a minimum if f'(x) changessign
from negativeto positive as x increasesthrough a. This meansthat f '(x) is
increasingat a so f"(a) > 0 .
The nature of a stationary point at x = b is a maximum if f '(x) changessign from
positiveto negativeas x increasesthrough b, that is f "(b) < 0 .
In summary,a stationarypoint c of f(x) is:
A local maximumif
f"(c) < 0 .
A local minimum if
f"(c) > 0 .
Note that if c is a stationarypoint and f"( c) = 0 , thenno informationaboutthe
stationarypoint canbe deduced.
Example11.13 Let f(x) = x3 + 3X2- 45x. Findthestationary
pointsof f. Usethe
secondderivative
testto findthenatureof thestationary
points.
Solution
f'(x) = 3X2 + 6x - 45
= 3(x2+ 2x -15)
= 3(x + 5)(x - 3) .
Thus f'(3) = 0 and f'(-5)
=0
so x = 3 and x = -5 arestationary
points.Also,
f "(x) = 6x + 6
f"(-5) = -24 < 0 and f"(3) = 24> 0
so x = - 5 is a local maximum,and x = 3 is a localminimum.
l
.'\
Differentiation3
The SecondDerivativeTest
.
Page10
.I
Example 11.14
Show that
f(x) = 2e-x-e-2x hasastationarypointatx=O and
use the secondderivativetest to determineits nature.
Solution
f'(x)
= -2e-x +2e-2x
=
fiCO)
-2
Also, f"(x)
+ 2 = 0 (since eo= 1) so x = 0 is a stationarypoint.
= 2e-x - 4e-2x and f"(O) = 2 - 4 = -2 < 0 .
Thus x = 0 is a localmaximum.
Find f'(x)
Example 11.15
and
f"(x) when f(x) = cos2x-2sinx. Show that x=7t/2
is a stationarypointand determineits nature.
Solution
f'(x)
= -2sin2x - 2cosx
f"(x)
= -4cos2x + 2sinx
Then
f'(7t/2) = -2sin7t - 2cos(7t/2) = 0 so 7t/2 is a stationarypoint.
f"(7t/2) = -:-4cos7t+ 2 sin(7t/2) = 4 + 2 = 6 > 0 so x = 7t/2 is
a localminimum.
!
Also
-
Exercises 11.6
1.
Find f '(x) and f "(x) when f(x) = e3x- 3ex. Showthat x = 0 is a
stationarypoint anddetennineits nature.
2.
Find f'(x) and f"(x) when f(x) = xe-x. Showthat x = 1 is a stationary
point anddetennineits nature.
3.
Find f'(x) and f"(x) when f(x) = X2 + cosx. Showthat x = 0 is a
stationarypoint anddetennineits nature.
...
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Differentiation3
FurtherExercises
':,.
to
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-"'---'-~.~-.c_c.-~
Further
Page11
,",
,~
Exercises
,:,:;.;
":
1.
If
y = (4x + 7)-1, showthat y' = -4y2 .
2.
If
y = 4cos3x, showthat y" + 9y = O.
3.
Let f(x) = ax2+ bx where a and b areconstants.If f"(2) = 6 and
f '(2) = 5, find a and b.
4.
Find f'(x) when (i) f(x) = x2cos3x
5.
Showthat f(x) = xcos2x satisfies
(ii) x5sin2x .
f"(x) + 4f(x) = Asin2x
where A is a constant,and find A.
6.
Find f '(x) when (i) f(x) =
~
(ii) f(x)
= -~~~
3x+l
7.
Let f(x) =
~~
.
5x +7x
. Expressf '(x) in the form A(2x + 9)-2, where A is a constant.
2x+9
8.
If
Y = 2ex- e-3x , showthat y" + 2y' - 3y = 0 .
9.
If
f(x) = exp(-5x2), showthat f'(x) = -10xf(x).
10. If
Y = xlnx find y" .
11. Showthat y = e5x satisfiesan equationof the form
y" - 3y' + Ay = 0
where A is a constant,andfind the valueof A. Showalsothat, with this value
of A, Y = e-2x alsosatisfiesthe equation.
12. Find f'(x) and f"(x) when f(x) = lnx - x. Showthat x = 1 is a stationary
point and determineits nature.
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Differentiation3
Answersto Exercises
Page12
-
-:
.
J -c
Answers to Exercises
"
I
Exercises 11.1
ill
1.
(i) 4x3-21x2
(ii)
3.
(i) 7(X2- x + 3)6(2x - 1)
Exercises 11.2
1. (i) -16cos2x
-15sin3x-cosx
(ii) -2x(X2 + 1)-2
(ii) 6x-4
Exercises 11.3
1. (i) 3x2cosx- X3sinx
3.
X(X2+ 1)-ln(x + 3) + (X2+ 1)ln
3.
A=6
3.
f'(x)
(iii)
-2x-3
I
3X2- 6X2(X3+ 1)-3
6
2.
sinx-xcosx
sin2x
= -13(5x - 1)-2, f"(x) = 130(5x- 1)-3
-
Exercises 11.5
1.
(i) -5e-sx + 3e3x (ii) -4xexp(-2x1
2.
e2x+ 2xe2x
3.
(i)
~
(ii)
x
!
[
j
;.
i
I
I
I
5.
~
3+x
A = -2
Exercises 11.6
1. local minimum
2.
local maximum
3.
local mimimum
Further Exercises
3.
4.
a=3, b=-7
(i) 2xcos3x - 3x2sin3x
5.
A = -4
. cosx(3x+l) - 3sinx
6
(1)
(3x+ 1)2
7
A = 42
10.
!
x
11. A=-10
12. local maximum
i"
(ii) 4(x + 2)s + 5(4x + 3)(x + 2)4
2.
Exercises 11.4
1. x2+2x-l
(x + 1)2
2.
(ii) 5x4sin2x+ 2xscos2x
.. 3~x2 +7x) - (3x+2Xl0x+7)
(11)
-
-
~X2 +7x)
-
=
-15x2 -20x-14
~X2 +7X)