chap 1 advanced differentiation
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WLB 10102 / WLB 10703 /LN/01 Chap 1 Advanced Differentiation
Mohd Nasir Mahmud@UniKL MICET Page 1 of 25 Wednesday, January 16, 2008
1.1 IMPLICIT DIFFERENTIATION
Procedure for Implicit Differentiation
Example 1
If )(xfy = is a differentiable function of x such that
yxyyx 232 32 +=+
find .dx
dy
Objectives: At the end of this lesson, the students should be able to:
i. differentiate the implicit function containing products and quotients. ii. find and evaluate a partial derivative (1st order) iii. determine the higher-order partial derivative (2nd order & mixed 2nd order) iv. verify that a function satisfies the given equation v. determine the total differential of the given function.
vi. use spital'oHl' rule to compute the indeterminate forms ∞∞
, 0
0 etc.
vii. apply the gradient function for optimization in the life sciences viii. find the rate of change of volume/area etc ix. determine the approximate error of a quantity caused by small changes in the
variables associated with the quantity.
CHAPTER 1 ADVANCED DIFFERENTIATION
Suppose an equation defines y implicitly as a differentiable function of x . To find
:dx
dy
Step 1. Differentiate both sides of the equation with respect to x . Remember that y is really a function of x for part of the curve and use the chain rule when differentiating terms containing y .
Step 2. Solve the differentiated equation algebraically for .dx
dy
WLB 10102 / WLB 10703 /LN/01 Chap 1 Advanced Differentiation
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Solution
Differentiate both sides of this equation term by term with respect to x :
( ) ( )yxdx
dyyx
dx
d232 32 +=+
( ) ( ) ( ) ( )ydx
dx
dx
dy
dx
dyx
dx
d232 32 +=+
( ) ( ) ( )dx
dyy
dx
dyx
dx
dyy
dx
dx 2)1(332
rulePower Extended
2
ruleProduct
22 +=
++4434421444 3444 21
dx
dy
dx
dyyxy
dx
dyx 2362 22 +=++
Finally, solve this equation for :dx
dy
( )
26
23
2326
22
22
−+−=∴
−=−+
yx
xy
dx
dy
xydx
dyyx
Example 2 Differentiate the following functions with respect to :x
a) 42y b) t3sin
Solution
a) Let 42yu = , then, by the function of a function rule:
dx
dyy
dx
dyy
dy
d
dx
dy
dy
du
dx
du
3
4
8
)2(
=
×=×=
b) Let tu 3sin= , then, by the function of a function rule:
dx
dtt
dx
dtt
dt
d
dx
dt
dt
du
dx
du
3cos3
)3(sin
=
×=×=
■
■
■
It is possible to differentiate an implicit function by using the function of a function rule, which may be
stated as dxdy
dydu
dxdu ×= .
i.e. f(y)u = .
WLB 10102 / WLB 10703 /LN/01 Chap 1 Advanced Differentiation MTH1134/LN/01 Chap 1 Advanced Differentiation
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Example 3 Differentiate the following functions with respect to :x
a) y5ln4 b) 2θ3
5
1 −e
Solution
a) Let yu 5ln4= , then, by the function of a function rule:
dx
dy
y
dx
dyy
dy
d
dx
dy
dy
du
dx
du
4
)5ln4(
=
×=×=
b) Let 2θ3
5
1 −= eu , then, by the function of a function rule:
dx
de
dx
de
d
d
dx
d
d
du
dx
du
θ
5
3
θ)
5
1(θ
θ
θ
2θ3
2θ3
−
−
=
×=×=
Differentiating implicit functions containing produ cts and quotients
Example 4 Determine ).2( 23 yxdx
d
Solution
In the product rule of differentiation let 32xu = and 2yv = . Thus
+=
+=
+
=
+=
ydx
dyxyx
yxdx
dyyx
xydx
dyyx
xdx
dyy
dx
dxyx
dx
d
322
64
)6)((2)2(
)2()()()2()2(
2
223
223
322323
■
■
■
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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Example 5 Find ).2
3(
x
y
dx
d
Solution
In the quotient rule of differentiation let yu 3= and xv 2= . Thus
−=
−=
−
=
−=
ydx
dyx
x
x
ydx
dyx
x
ydx
dyx
x
xdx
dyy
dx
dx
x
y
dx
d
2
2
2
2
2
34
66
4
)2)(3(3)2(
)2(
)2()3()3()2()
2
3(
Do it yourself
1. In Problems (a) – (e) differentiate the given functions with respect to x .
a) 53y
b) θ4cos2
c) k
d) t3ln2
5
e) 12
4
3 +ye
f) y3tan2
2. Differentiate the following with respect to y .
a) θ2sin3
b) 34 x c)
te
2
3. Differentiate the following with respect to u .
a) )13(
2
+x
b) θ2sec3
c) 3
2
y
4. Determine
a) )3( 32 yxdx
d
b) )4
3(
v
u
du
d
c) dy
dz given yxz ln2 3=
5. Find )5
2(
x
y
dx
d
■
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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1.2 PARTIAL DIFFERENTIATION
1.2.1 FIRST ORDER PARTIAL DERIVATIVES Example 6 If ,),( 223 yxyxyxf += find
a) xf b) yf
Solution a) For xf , hold y constant and find the derivative with respect to :x
22 23),( xyyxyxf x += ■
b) For yf , hold x constant and find the derivative with respect to :y
yxxyxf y23 2),( += ■
Example 7 Finding and evaluating a partial derivative Let ).3sin( 32 yxxz += Evaluate:
a) )0,
3
π(x
z
∂∂
b) yz at (1, 1).
Solution
a) )3)(3cos()3sin(2 323 yxxyxxx
z +++=∂∂
)3cos(3)3sin(2 323 yxxyxx +++= Thus,
=∂∂
)0,3
π(x
z
3
π)1(
3
π)0(
3
π2πcos
3
π3πsin
3
π2
222
−=−+=
+
■
For ),,( yxfz = the partial derivatives and yx ff are denoted by
)(),(),( fDzyxfxx
z
x
fyxf xxx ==
∂∂=
∂∂=
∂∂=
and
)(),(),( fDzyxfyy
z
y
fyxf yyy ==
∂∂=
∂∂=
∂∂=
The values of the partial derivatives of ),( yxf at the point ),( ba are denoted by
),(),(
bafx
fx
ba
=∂∂
and ),(),(
bafy
fy
ba
=∂∂
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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b) )3cos(3)3)(3cos( 322232 yxyxyyxxzy +=+= so that
4cos3)13cos()1()1(3)1,1( 22 =+=yz ■
Example 8 Partial derivative of a function of three variables Let ;2),,( 322 yzxyxzyxf ++= determine:
a) xf b) yf c) zf
Solution a) For xf , think of f as a function of x alone with y and z treated as constants:
222),,( yxzyxf x += ■
b) 34),,( zxyzyxf x += ■
c) 23),,( yzzyxf z = ■ Example 9 Partial derivative of an implicitly defined function Let z be defined implicitly as a function of x and y by the equation
xyzzx =+ 32
Determine x
z
∂∂
and y
z
∂∂
.
Solution Differentiate implicitly with respect to x , treating y as a constant:
132 22 =∂∂+
∂∂+
x
zyz
x
zxxz
Then solve the equation for x
z
∂∂
:
22 3
21
yzx
xz
x
z
+−=
∂∂
Similarly, holding x constant and differentiating implicitly with respect to y , we find
03 232 =∂∂++
∂∂
y
zyzz
y
zx
so that
22
3
3yzx
z
y
z
+−=
∂∂
■
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
Mohd Nasir Mahmud@UniKL MICET Page 7 of 25 Wednesday, January 16, 2008
Do it yourself
1. If yyxxz 325 234 −+= find
a) x
z
∂∂
and b) .y
z
∂∂
2. Given ,2cos3sin4 txy = find
a) x
y
∂∂
and b) .t
y
∂∂
3. If xyz sin= show that =∂∂x
z
y
1.
1
y
z
x ∂∂
4. In Problems (a) – (f), find x
z
∂∂
and .y
z
∂∂
a) xyz 2=
b) 23 2 yxyxz +−=
c) y
xz =
d) )34sin( yxz +=
e) yx
yyxz
12
23 +−=
f) yxz 4sin3cos=
1.2.2 HIGHER-ORDER PARTIAL DERIVATIVES (2ND ORDER & MIXED 2 ND ORDER)
Given ).,( yxfz = 2nd order partial derivatives
xxxx ffx
f
xx
f ==
∂∂
∂∂=
∂∂
)(2
2
yyyy ffy
f
yy
f ==
∂∂
∂∂=
∂∂
)(2
2
Mixed 2nd order partial derivatives
yxxy ffy
f
xyx
f ==
∂∂
∂∂=
∂∂∂
)(2
xyyx ffx
f
yxy
f ==
∂∂
∂∂=
∂∂∂
)(2
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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Example 10 Higher-order partial derivatives of a function of two variables For ,325),( 32 yxyxyxfz +−== determine these higher-order partial derivatives.
a) yx
z
∂∂∂ 2
b) xy
f
∂∂∂ 2
c) 2
2
x
z
∂∂
d) )2 ,3(xyf
Solution a) First differentiate with respect to y ; then differentiate with respect to .x
292 yxy
z +−=∂∂
( ) 292 22
−=+−∂∂=
∂∂
∂∂=
∂∂∂
yxxy
z
xyx
z ■
b) Differentiate first with respect to xand then differentiate with respect to :y
yxx
z210 −=
∂∂
( ) 22102
−=−∂∂=
∂∂
∂∂=
∂∂∂
yxyx
z
yxy
z ■
c) Differentiate first with respect to x twice:
( ) 102102
2
=−∂∂=
∂∂
∂∂=
∂∂
yxxx
z
xx
z ■
d) Evaluate the mixed partial found in part (b) at the point (3, 2):
2)2 ,3( −=xyf ■
Example 11 Partial derivatives of functions of two variables Determine ,,, xxyxxy fff and ,xxyf where yyexxf 2)y ,( = .
Solution We have the partial derivatives
yx xyef 2= yy
y yexexf 22 +=
The mixed partial derivatives (which must be the same by the previous theorem) are
yyyxxy xyexeff 22)( +== ■ yy
xyyx xyexeff 22)( +== ■
Finally, we compute the second- and higher-order partial derivatives:
yxxxx yeff 2)( == ■ and yy
yxxxxy yeeff 22)( +== ■
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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Example 12 Verifying that a function satisfies the heat equation
Verify that c
xetxT t cos) ,( −= satisfies the heat equation, .
2
22
x
Tc
t
T
∂∂=
∂∂
Solution
c
xe
t
T t cos−−=∂∂
and
c
xe
c
c
xe
cxx
T
t
t
cos1
sin1
2
2
2
−
−
−=
−∂∂=
∂∂
Thus, T satisfies the heat equation 2
22
x
Tc
t
T
∂∂=
∂∂
■
1.3 TOTAL DIFFERENTIAL
Example 13 Determine the total differential of the given functions:
a) zyxzyxf 652),,( 43 −+= b) )23ln(),( 22 xyxyxf −= Solution
a) dzdyydxxdzz
fdy
y
fdx
x
fdf 6206 32 −+=
∂∂+
∂∂+
∂∂= ■
b) dyy
fdx
x
fdf
∂∂+
∂∂=
dyxy
yxdx
xy
xxyx
dyxy
yxdx
xyxxyx
23
6
23
2)23ln(2
23
6
23
2)23ln(2
2
2
2
22
22
222
−+
−−−=
−+
−−+−=
■
The total differential of the function ),( yxf is
dyyxfdxyxfdyy
fdx
x
fdf yx ),(),( +=
∂∂+
∂∂=
where dx and dyare independent variables. Similarly, for a function of three variables
),,( zyxfw = the total differential is
dzz
fdy
y
fdx
x
fdf
∂∂+
∂∂+
∂∂= ( أ ).……………
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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Example 14 If ),( yxfz = and ,1232 ++=y
xyxz determine the total differential, .dz
Solution The total differential is the sum of the partial differentials, i.e.
222
3
23
22
y
xyx
y
z
yxy
x
z
dyy
zdx
x
zdz
−=∂∂
+=∂∂
∂∂+
∂∂=
Hence dyy
xyxdx
yxydz
−+
+=
2223 2
32
2 ■
Example 15 If ),,( wvufz = and 232 423 vwvuz +−= find the total differential, .dz Solution The total differential
22
3
12
82
6
vww
z
vwv
z
uu
z
dww
zdv
v
zdu
u
zdz
=∂∂
+−=∂∂
=∂∂
∂∂+
∂∂+
∂∂=
Hence dwwvdvvwududz )12()28(6 223 +−+= ■
Do it yourself
1. In Problems (a) – (f), find the total differential .dz
a) 23 yxz +=
b) xxyz cos2 −=
c) yx
yxz
+−=
d) yxz ln=
e) 4−+=y
xxyz
2. If ),,( cbafz = and ,32 2 abccbabz +−= find the total differential .dz
(i.e. y is kept constant)
(i.e. x is kept constant)
(i.e. wv and are kept constant)
(i.e. wu and are kept constant)
(i.e. vand u are kept constant)
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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1.4 L’HOPITAL’S RULE
Example 17 Using l’Hopital’s rule to compute a familiar trigonometric limit
Evaluate .sin
lim0 x
xx→
Solution Note that this is an indeterminate form because sin x and x both approach 0 as .0→x This means that l’Hopital’s rule applies:
11
1
1
coslim
)(
)(sinlim
sinlim
000====
→→→
x
xdx
d
xdx
d
x
xxxx
■
Example 18 l’Hopital’s rule with a 0/0 form
Evaluate .8
128lim
3
7
2 −−
→ x
xx
Solution
For this example, 128)( 7 −= xxf and 8)( 3 −= xxg , and the form is 0/0.
)8(
)128(lim
8
128lim
3
7
23
7
2−
−=
−−
→→x
dx
d
xdx
d
x
xxx
Let f and g be differentiable functions with 0)( ≠′ xg on an open interval containing
c (except) possibly at c itself). Suppose )(
)(lim
xg
xfcx→
produces an indeterminate form 0
0 or
∞∞
and that
Lxg
xfcx
=′′
→ )(
)(lim
where L is either a finite number, ,∞+ or .∞− Then
Lxg
xfcx
=→ )(
)(lim
The theorem also applies to one-sided limits and to limits at infinity (where +∞→x and −∞→x ).
l’Hopital’s rule
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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3
112
3
)2(7
3
7lim
3
7lim
4
4
2
2
6
2
==
=
=
→
→
x
x
x
x
x
Example 19 Limit is not an indeterminate form
Evaluate .sec
cos1lim
0 x
xx
−→
Solution You must always remember to check that you have an indeterminate form before applying l’Hopital’s rule. The limit is
01
0
seclim
)cos1(lim
sec
cos1lim
0
0
0==
−=−
→
→
→ x
x
x
x
x
x
x ■
ATTENTION: If you blindly apply l’Hopital’s rule in Example 19, you obtain the WRONG answer:
xx
x
x
xxx tansec
sinlim
sec
cos1lim
00 →→=−
11
1
sec
coslim
0===
→ x
xx
Example 20 l’Hopital’s rule applied more than once
Evaluate .sin
lim30 x
xxx
−→
Solution This is a 0/0 indeterminate form, and we find that
2030 3
cos1lim
sinlim
x
x
x
xxxx
−=−→→
This is still the indeterminate form 0/0, so l’Hopital’s rule can be applied once again:
6
1)1(
6
1sinlim
6
1
6
)sin(lim
3
cos1lim
0020===−−=−
→→→ x
x
x
x
x
xxxx
■
Example 21 l’Hopital’s rule with an ∞∞ form
Evaluate .253
132lim
2
2
−++−
+∞→ xx
xxx
Solution
We could compute this limit by multiplying by ( ) ( )22 11 xx . Instead, we note that this is of
the form ∞∞ and apply l’Hopital’s rule:
56
34lim
253
132lim
2
2
+−=
−++−
+∞→+∞→ x
x
xx
xxxx
■
Simplify
Limit of a quotient
Apply l’Hopital’s rule again
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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3
2
6
4lim ==
+∞→x ■
Example 22 Using l’Hopital’s rule with other limit properties
Evaluate .cos
4sin)cos1(lim
30 xx
xxx
−→
Solution The limit has the form 0/0, but direct application of l’Hopital’s rule leads to a real mess. Instead, we compute the given limit by using the product rule for limits first, followed by two simple applications of l’Hopital’s rule. Specifically, using the product rule for limits, we have
−=−→→→→ xx
x
x
x
xx
xxxxxx cos
1lim
4sinlim
)cos1(lim
cos
4sin)cos1(lim
002030
( )( ) 2142
1
1
1
1
)1(4
2
coslim
cos
1lim
1
4cos4lim
2
sinlim
0
000
=
=
=
=
→
→→→
x
x
x
x
x
x
xxx
Example 23 Hypothesis of l’Hopital’s rule are not satisfied
Evaluate .cos
sinlim
xx
xxx −
++∞→
Solution This limit has the indeterminate form ∞∞ . If you try to apply l’Hopital’s rule, you find
x
x
xx
xxxx sin1
cos1lim
cos
sinlim
++=
−+
+∞→+∞→.
The limit on the right does not exist, because both sin x and cos x oscillate between -1 and 1
as .+∞→x Recall that l’Hopital’s rule applies only if Lxg
xfcx
=′′
→ )(
)(lim or is .∞± This does
not mean that the limit of the original expression does not exist or that we cannot find it; it simply means that we cannot apply l’Hopital’s rule. To find this limit, factor out an x from the numerator and denominator and proceed as follows:
101
01cos
1
sin1
limcos
1
sin1
limcos
sinlim =
−+=
−
+=
−
+=
−+
+∞→+∞→+∞→
x
xx
x
x
xx
x
xx
xx
xxxxx
■
OTHER INDETERMINATE FORMS
Example 24 Limit of the form ∞1
■
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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Show that .1
1lim ex
x
x=
++∞→
Solution
Note that this limit is indeed of the indeterminate form ∞1 . Let
x
x xL
+=+∞→
11lim
Take the logarithm of both sides:
x
x
xx
x
xL
x
x
x
x
x
x
1
11ln
lim
11lnlim
11lnlim
11limlnln
+=
+=
+=
+=
+∞→
+∞→
+∞→
+∞→
101
1
11
1lim
1
1
11
1
lim
2
2
=+
=
+=
−
−+=
+∞→
+∞→
x
x
xx
x
x
Thus, 1ln =L and eeL == 1 ■ Example 25 l’Hopital’s rule with the form ∞⋅0
Evaluate .tan2
πlim
2πxx
x
−−→
Solution The limit has the form ∞⋅0 , because
+∞==
−−− →→
xxxx
tanlim and 02
πlim
2π2π
ln x is continuous
Property of logarithms
Form 0/0
l’Hopital’s rule
Simplify
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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Write x
xcot
1tan = to obtain
x
xxx
xx cot2
π
limtan2
πlim
2π2π
−=
−−− →→
1)sin(lim
csc
1lim
2
2π
22π
−=−=−
=
−
−
→
→
x
x
x
x
Example 26 Limit of the form 00
Find x
xxsin
0lim
+→.
Solution
This is a 00 indeterminate form. We begin by using properties of logarithms.
x
x
xx
x
xL
xL
x
x
x
x
x
x
x
x
cscln
lim
ln)(sinlim
lnlim
limlnln
lim
0
0
sin
0
sin
0
sin
0
+
+
+
+
+
→
→
→
→
→
=
=
=
=
=
0)0)(1(
cossinsin
lim
cossin
lim
cotcsc1
lim
0
2
0
0
==
−
=
−=
−=
+
+
+
→
→
→
x
x
x
xxx
x
xx
x
x
x
x
Thus, 10 == eL ■
Example 27 Limit of the form 0∞ Find x
xx1lim
+∞→.
Solution
This is a limit of the indeterminate form 0∞ .
If ,lim 1 x
xxL
+∞→= then
Form 0/0
l’Hopital’s rule
■
ln x is continuous This is ∞.0 form
This is ∞∞ form
l’Hopital’s rule
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
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01
1
lim
lnlim
ln1
lim
lnlim
limlnln
1
1
=
=
=
=
=
=
+∞→
+∞→
+∞→
+∞→
+∞→
x
x
x
xx
x
xL
x
x
x
x
x
x
x
Thus, we have ;0ln =L therefore, 10 == eL ■ Example 28 l’Hopital’s rule with the form ∞−∞
Evaluate
−+→ xxx sin
11lim
0.
Solution
As it stands, this has the form ∞−∞ , because +∞→x
1 and +∞→
xsin
1 as 0→x from
the right. However, using a little algebra, we find
xx
xx
xx xx sin
sinlim
sin
11lim
00
−=
−++ →→
This limit is now of the form 0/0, so the hypotheses of l’Hopital’s rule are satisfied. Thus,
xxx
x
xx
xxxx cossin
1coslim
sin
sinlim
00 +−=−
++ →→
02
0
cos)sin(cos
sinlim
0==
+−+−=
+→ xxxx
xx
■
ATTENTION: Not all limits that appear indeterminate actually are indeterminate. For example,
form 0
0ln
tanlim
form )(-- )ln(csclim
form 0 0)(sinlim
0
0
1
0
∞=
∞∞++∞=−
=
+
+
+
→
→
∞
→
x
x
xx
x
x
x
x
x
Other such “false indeterminate forms” include ,0 ),( ∞+∞+∞+ and ,.∞∞ which are all actually infinite.
Do it yourself
1. An incorrect use of l’Hopital’s rule is illustrated in the following limit computations. In each case, explain what is wrong and find the correct value of the limit.
The limit of a log is the log of the limit
This is ∞∞
l’Hopital’s rule
This is 0/0 form
WLB 10102 / WLB 10703 / SN/01 Chap 1 Advanced Differentiation
Mohd Nasir Mahmud@UniKL MICET Page 17 of 25 Wednesday, January 16, 2008
a) 01
sinlim
cos1lim
ππ==−
→→
x
x
xxx
b) 01
coslim
sinlim
2π2π==
→→
x
x
xxx
2. Find each of the limits in Problems (a) – (g).
a) 1
1lim
2
3
1 −−
→ x
xx
b) 1
1lim
10
1 −−
→ x
xx
c) 67
43lim
2
2
1 +−−+
→ xx
xxx
d) sin
cos1lim
3
2
0 x
xx
−→
e) xx
xxx −
−→ tan
sinlim
0
f) 20
sinlim
x
xxx
−→
g) x
xx tan2
sec3lim
2π +→
h)
+−−→
xxx
tan2π
1lim
)2π(
i) 2
11lim
3x
++∞→ xx
j) )ln(ln
limx
xx +∞→
k) ( ) limx1
0xex
x+
+→
l) ln1
lim20
−+→
xxx
1.5 APPLICATION OF DIFFERENTIATION
1.5.1 GRADIENT FUNCTIONS (OPTIMIZATION IN THE LIFE SCIENCES)
Example 29 Maximization Applied to Enzymes An enzyme is a protein that acts as a catalyst for increasing the rate of a chemical reaction that occurs in cells. In a certain reaction, an enzyme is converted to another enzyme called the product. The product acts as a catalyst for its own formation. The rate R at which the product is formed (with respect to time) is given by
),( plkpR −= where l is the total initial amount of both enzymes, p is the amount of the product enzyme, and k is a positive constant. For what value of p will Rbe a maximum? Solution We can write ).( 2pplkR −= Setting 0=dpdR and solving for p gives
.2
,0)2(l
pplkdp
dR =∴=−=
Now, .222 kdpRd −= Since ,0>k the second derivative is always negative. Hence,
2lp = gives a relative maximum. Moreover, since R is a continuous function of ,p
we conclude that we indeed have an absolute maximum when 2lp = ■ Example 30 Modeling Problem: Maximum concentration of a drug Let )(tC denote the concentration in the blood at time t of a drug injected into the body intramuscularly. In a classic paper by E. Heinz (“Probleme bei der Diffusion
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kleiner Substanzmengen innerhalb des menschlichenrpersoK && ”, Biochem., Vol. 319 (1949), pp. 482-492), it was observed that the concentration may be modeled by
0 t)()( ≥−−
= −− btat eeab
ktC
where ba , (with ),ab > and k are positive constants that depend on the drug. At what time does the largest concentration occur? What happens to the concentration as
?+∞→t Solution To locate the extrema, we solve .0)( =′ tC
[ ] )()()(
)()(
atbtbtat
btat
aebeab
kebea
ab
k
eeab
k
dt
dtC
−−−−
−−
−−
=−−−−
=
−−
=′
We see that 0)( =′ tC when
atbtatbt
atbt
eeea
b
aebe
−−
−−
==
=
a
batbt ln=−
a
b
abt ln
1
−= ■
( )0
00
1lim
1lim
)(lim)(lim
=
−−
=
−−
=
−−
=
+∞→+∞→
−−
+∞→+∞→
ab
k
eeab
k
eeab
ktC
bttatt
btat
tt
Example 31 The production of blood cells plays an important role in medical research involving leukemia and other so-called dynamical diseases. In 1977, a mathematical model was developed by A. Lasota that involved the cell production function
rsxseAxxP −=)( where rsA and ,, are positive constants and x is the number of granulocytes (a type of white blood cell) present. Find the granulocytes level x that maximizes the production function P .
■
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Solution
rx
rx
rx
x
xr
x
esAxexr
sA
sAxeer
sAxxP
eAxxP
ss
s
s
ss
rsxsrsxs
srsxrsxs
rsxs
=∴=
=
=
=
=+
−=′
=
+−
−
−
−−−
−−−
−
1
1
1
1
1 0.)(
)(
Example 32 Beehives are formed by packing together cells that may be modeled as regular hexagonal prisms open at one end. It can be shown that a cell with hexagonal side of length s and prism height h has surface area
)θcsc3θcot(5.16)θ( 2 +−+= sshS
for .2
πθ0 << What is the angle θ (to the nearest degree) that minimizes the surface
area of the cell (assuming that s and h are fixed)? Solution
oo
sS
sshS
557.54θ
3
1θcos
θsin
θcos3
θsin
1
θcot3θcsc
θcotθcsc3θcsc
0)θcotθcsc3θcsc(5.1)θ(
)θcsc3θcot(5.16)θ(
2
22
2
≈=∴
=
=
=
=
=−−=′
+−+=
1.5.2 RATES OF CHANGE Sometimes it is necessary to solve problems in which different quantities have different rates of change. From equation (the total differential)
dzz
fdy
y
fdx
x
fdf
∂∂+
∂∂+
∂∂= ,
the rate of change of f , dt
df is given by:
■
■
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Example 33 The height of a right circular cone is increasing at 3 mm/s and its radius is decreasing at 2 mm/s. Determine, correct to 3 significant figures, the rate at which the volume is changing (in cm3/s) when the height is 3.2 cm and the radius is 1.5 cm. Solution
Volume of a right circular cone, hrV 2π
3
1=
Using equation (ب), the rate of change of volume,
2π
3
1 and π
3
2r
h
Vrh
r
Vdt
dh
h
V
dt
dr
r
V
dt
dV
=∂∂=
∂∂
∂∂+
∂∂=
Since the height is increasing at 3 mm/s, i.e. 0.3 cm/s, then 3.0+=dt
dh
and since the radius is decreasing at 2 mm/s, i.e 0.2 cm/s, then 2.0−=dt
dr.
Hence
( ) ( )
2
2
π1.0π3
4.0
3.0π3
12.0π
3
2
rrh
rrhdt
dV
+−=
+
+−
=
However, h = 3.2 cm and r = 1.5 cm. Hence
scm
dt
dV
/ 304.1
)5.1(π)1.0()2.3)(5.1(π3
4.0
3
2
−=
+−=
Thus the rate of change of volume is 1.304 cm3/s decreasing.
Example 34 The area A of a triangle is given by ,sin2
1BacA = where B is the
angle between sides a and c. If a is increasing at 0.4 units/s, c is decreasing at 0.8 units/s and B is increasing at 0.2 units/s, find the rate of change of the area of the triangle, correct to 3 significant figures, when a is 3 units, c is 4 units and B is 6π radians. Solution Using equation(ب), the rate of change of area,
dt
dz
z
f
dt
dy
y
f
dt
dx
x
f
dt
df
∂∂+
∂∂+
∂∂= (ب)……
■
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Mohd Nasir Mahmud@UniKL MICET Page 21 of 25 Wednesday, January 16, 2008
dt
dB
B
A
dt
dc
c
A
dt
da
a
A
dt
dA
∂∂+
∂∂+
∂∂=
Since Bca
ABacA sin
2
1 ,sin
2
1 =∂∂= ,
units/s 2.0 and units/s 8.0 units/s, 4.0
cos2
1 and sin
2
1
=−==
=∂∂=
∂∂
dt
dB
dt
dc
dt
da
BacB
ABa
c
A
Hence
)2.0(cos2
1)8.0(sin
2
1)4.0(sin
2
1
+−
+
= BacBaBcdt
dA
When a = 3, c = 4 and B = 6
π then:
s
dt
dA
/units 839.0
)2.0(6
πcos)4)(3(
2
1)8.0(
6
πsin)3(
2
1)4.0(
6
πsin)4(
2
1
2=
+−
+
=
1.5.3 SMALL INCREMENT AND APPROXIMATION It is often useful to find an approximate value for the change (or error) of a quantity caused by small changes (or errors) in the variables associated with the quantity. If ),,( zyxfw = and
zyx δ ,δ ,δ denote small changes in zyx ,, respectively, then the corresponding approximate change fδ in f is obtained from equation ( أ ) by replacing the differentials by the small changes. Thus Example 35 Pressure p and volume V of a gas are connected by the equation
k pV =4.1 . Determine the approximate percentage error in k when the pressure is increased by 4% and the volume is decreased by 1.5%. Solution Using equation ( � ), the approximate error in k,
VV
kp
p
kk δδδ
∂∂+
∂∂≈
Let p, V and k refer to the initial values.
Since 4.04.14.1 4.1 and then VV
kV
p
k pVk =
∂∂=
∂∂= .
Since the pressure is increased by 4%, the change in pressure .04.0100
4δ ppp =×=
■ (correct to 3 significant figures)
zz
fy
y
fx
x
ff δδδδ
∂∂+
∂∂+
∂∂≈ …………….( � )
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Since the volume is decreased by 1.5%, the change in volume
.015.0100
5.1δ VVV −=×−=
Hence the approximate error in k,
[ ].
100
9.1
100
9.1019.0
)015.0(4.104.0
)015.0)(4.1()04.0()(δ
4.14.1
4.1
4.04.1
kpVpV
pV
VpVpVk
≈≈≈
−≈
−+≈
i.e. the approximate error in k is a 1.9% increase. Example 36 Modulus of rigidity ,)θ( 4 LRG = where R is the radius, θ the angle of twist and L the length. Determine the approximate percentage error in G when R is increased by 2%, θ is reduced by 5% and L is increased by 4%. Solution
Using LL
GGR
R
GG δδθ
θδδ
∂∂+
∂∂+
∂∂≈ .
Since .θ
and θ
,θ4
,)θ(
2
4434
L
R
L
G
L
RG
L
R
R
G
L
RG
−=∂∂=
∂∂=
∂∂=
Since R is increased by 2%, RRR 02.0100
2δ == . Similarly,
.04.0Lδ and θ05.0δθ L=−= Hence
( ) ( )
GGL
R
LL
R
L
RR
L
RG
100
1δ i.e. ,
θ01.0
04.0θ
θ05.0)02.0(θ4
δ
4
2
443
−≈−≈
−+−
+
≈
Hence the approximate percentage error in G a 1% decrease.
1.5.3 DIRECTIONAL DERIVATIVES AND THE GRADIENT
Definition
Let f be a function of two variables, and let juiuu 21 += be a unit vector. The
directional derivative of f at ) ,( ooo yxP in the direction of u is given by
h
yxfhuyhuxfyxfD oooo
hoou
) ,() ,(lim) ,( 21
0
−++=→
provided the limit exists.
■
■
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Example 37 Find the directional derivative of 3223),( yxyxf +−= at the point
)2 ,1(P in the direction of the unit vector jiu23
21 −= .
Solution
First, find the partial derivatives xyxf x 4) ,( −= and .3) ,( 2yyxf y = Then since
21
1 =u and 23
2 −=u , we have
4.123622
3)2(3
2
1)1(4
23
)2 ,1(21
)2 ,1()2 ,1(
2 −≈−−=
−+
−=
−+
= yxu fffD
The Gradient
Let f be a differentiable function at ),( yx have partial derivatives ) ,( yxfx and
). ,( yxf y Then the gradient of f , denoted by f∇ (pronounced “del eff”), is a vector
given by
jyxfiyxfyxf yx ) ,() ,() ,( +=∇
The value of the gradient at the point ) ,( ooo yxP is denoted by
jyxfiyxff ooyooxo ) ,() ,( +=∇
Example 38 Find ) ,( yxf∇ for 32),( yyxyxf +=
Solution
Begin with the partial derivatives:
xyyyxx
yxfx 2)() ,( 32 =+∂∂= and 2232 3)() ,( yxyyx
yyxf y +=+
∂∂=
Then jyxixyyxf )3(2) ,( 22 ++=∇ ■
■
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Example 39 Find the directional derivative of )ln(),( 32 yxyxf += at )3- ,1(oP in the
direction of .32 jiv −=
Solution
32
2) ,(
yx
xyxfx +
= , so 262
)3- ,1( −=xf
32
23) ,(
yx
yyxf y +
= , so 2627
)3- ,1( −=yf
jiffo 2627
262
)3,1( −−=−∇=∇
A unit vector in the direction of v is
)32(13
1
)3(2
3222
jiji
v
vu −=
−+
−==
Thus,
3381377
13
3
26
27
13
2
26
2.) ,(
=
−
−+
−=∇= ufyxDu
Example 40 Heat flow application
The set of points ),( yx with 50 ≤≤ x and 50 ≤≤ y is a square in the first quadrant
of the xy-plane. Suppose this square is heated in such a way that 22),( yxyxT += is the temperature at the point ).,( yxP In what direction will heat flow from the point
?)4 ,3(oP
Solution
The heat flow TkH ∇−=ˆ where =k the thermal conductivity (a positive constant)
From 22),( yxyxT += , we have jyixT 22 +=∇ .
oT∇ is the gradient at oP , therefore .86 jiTo +=∇
Thus, the heat flow at oP satisfies )86(ˆ jikTkH oo +−=∇−= .
■
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Mohd Nasir Mahmud@UniKL MICET Page 25 of 25 Wednesday, January 16, 2008
Because the thermal conductivity k is positive, we can say that heat flows from oP in
the direction of the unit vector u given by
jiji
u5
4
5
3
)8()6(
)86(22
−−=−+−
+−= ■