Ÿ Differentiate with respect to x y HxL = I2 x3 + 5M I7 x4 + 3M
1.
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By using the formula for the derivative of a product we obtain
âu
uHxL = 2 x3 +5,
= 6 x2
âx âv
Out[16]=
vHxL = 7x4 +3,
= 28 x3
âx ây âx
= 28 H2 x3 + 5L x3 + 6 H7 x4 + 3L x2
ây or
âx
= 2 x2 H49 x4 + 70 x + 9L
2 ãx yHxL =
2.
5
x
Tick to show solution
By using the quotient rule we obtain
âu
uHxL = 2ã x ,
vHxL =
Out[48]=
= 2 ãx
âx âv
5
=
x,
âx ây =
âx
2 ãx 5 x4•5
ây =
or
âx 3.
+ 2 ãx
5
1 5 x4•5
x
2 ã x H5 x+1L 5 x4•5
yHxL = lnHxL I12 x3  4M
2  Numerical Integration
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By using the product rule we obtain
âu uHxL = ln x,
=
âx âv
Out[17]=
vHxL =12x3 4,
1 x
= 36 x2
âx ây =
12 x3 4 x
âx
+ 36 x2 logHxL
f HxL = Ix2 + 3M
2
4.
Tick to show solution
By using the chain rule we get
âu
uHxL = x2 + 3,
= 2x
âx
Out[33]=
f HuL = u2 ,
âf âx
âf = 2u
âu
= 4 x Hx2 + 3L
6 x2  4 x + 9
5.
yHxL =
ln HxL
Tick to show solution
By using the quotient rule we obtain
âu
uHxL = 6x2 4x+9,
= 12 x  4
âx
Out[13]=
âv vHxL = lnHxL,
=
âx ây
1
=
âx 6.
HlnHxLL2
1 x
H12 x  4L ln HxL  I6 x2  4 x + 9M x1 =
y HxL = H2 x + 4L4
12 x4 logHxL

6 x2 4 x+9 x log2 HxL
Numerical Integration  3
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By using the chain rule we get
âu uHxL = 2x + 4,
=2
âx
Out[32]=
f HuL = u4 ,
âf âx
âf
= 4 u3
âu
= 8 H2 x + 4L3
y HxL = H7 6 xL2
7.
Tick to show solution
By using the chain rule we get
âu uHxL =7  6x, f HuL = u2 ,
âf âx 8.
= 6
âx âf
Out[31]=
= 2u
âu
= 12 H7  6 xL 3
yHxL = 7 ãx
x
Tick to show solution
By using the product rule we get
âu
uHxL = 7ã x , Out[19]=
vHxL =
= 7 ãx
âx âv
3
=
x,
âx ây =
âx ây
=
or
âx
9.
7 ãx 3 x2•3
+ 7 ãx
3
1 3 x2•3
x
7 ã x H3 x+1L 3 x2•3
y HxL =
2 x5 x+1
4  Numerical Integration
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By using the formula for the derivative of a quotient we obtain
âu
uHxL = 2x5 ,
= 10 x4
âx âv
Out[8]=
vHxL = x+1,
=1
âx ây =
10 x4 Hx + 1L  2 x5 Hx + 1L
=
2
âx
10 x4 x+1

2 x5
Hx+1L2
yHxL = Ix2 + 3 x + 4M
10.
5
Tick to show solution
By using the chain rule we get
âu
uHxL = x2 + 3x + 4,
= 2x+3
âx
Out[29]=
f HuL = u5 ,
âf âx
âf
= 5 u4
âu
= 5 H2 x + 3L Hx2 + 3 x + 4L4
y HxL =
11.
x3  2 x2 + 1
Tick to show solution
By using the quotient rule we obtain
âu
uHxL = x3 2,
= 3 x2
âx âv
Out[7]=
vHxL = x2 +1,
= 2x
âx ây =
âx
12.
3 x2 Ix2 + 1M  Ix3  2M 2 x Ix2 + 1M
2
y HxL =
2 x2  5 3
x +8
=
3 x2 x2 +1

2 x Ix3 2M Ix2 +1M
2
Numerical Integration  5
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By using the quotient rule we obtain
âu
uHxL = 2 x2 5,
= 4x
âx âv vHxL = 3 x +8,
Out[9]=
2
âx ây
4x
=
âx ây
2 J3
âx
13.
3 I2 x2 5M
2 J3
2
x +8N
x
x
64 x3•2 +18 x2 +15
=
or

x +8
3
3
=
yHxL =
2
x +8N
x
6x+5
Tick to show solution
By using the chain rule we get
âu uHxL = 6x+5,
=6
âx
Out[25]=
f HuL = u1•2 ,
âf
âf
1
=
âu
2
u
3
=
6 x+5
âx
lnHxL 14.
yHxL =
5 x3  4
Tick to show solution
By using the quotient rule we obtain
âu uHxL = ln x, Out[14]=
vHxL = 5x3 4,
=
âx âv
1 x
= 15 x2
âx ây =
âx
15.
x1 I5 x3  4M  ln HxL 15 x2 I5 x3  4M
2
yHxL = H3 x + 4L2
=
1
x I5 x3 4M

15 x2 logHxL I5 x3 4M
2
6  Numerical Integration
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By using the chain rule we get
âu uHxL = 3x + 4,
=3
âx
Out[28]=
f HuL = u2 ,
âf =
âu
âf = 
âx
2 u3
6
H3 x+4L3
3 yHxL =
16.
3
2
x 5
x
Tick to show solution
By using the quotient rule we obtain
âu uHxL = 3,
=0
âx âv
3
vHxL = 2 x 5 x ,
=
âx
2 3 x2•3
5
2
x
Out[49]= 2
3
ây = 
âx
K2
3 x2•3 3
ây
x
x O
2
2 K25
âx
2
x 5
15
=
or
5

6
x 4 6
x O x4•3 2
If we rewrite yHxL = 3 J2
3
x 5 x N
yHxL = H1  8 xL3
17.
Tick to show solution
By using the chain rule we get
âu uHxL = 1  8x,
= 8
âx
Out[27]=
f HuL = u3 ,
âf =
âx
24
âf
H18 xL4
=
âu
3 u4
1
then the chain rule can be used.
Numerical Integration  7
18.
ã x I4 x  2 x3 + 3M
yHxL =
Tick to show solution
By using the product and chain rules we obtain
âu
ãx ,
uHxL =
ãx
=
2
âx
âv
Out[37]=
vHxL = 4x  2x3 + 3,
= 4  6 x2
âx ây
1
=
2
âx ây
=
or
âx
ã x H2 x3 + 4 x + 3L +
ã x H2 x3  12 x2 + 4 x + 11L
1 2
1
19.
ã x H4  6 x2 L
yHxL = lnHsinHxLL 2
cos2 H4 xL
Tick to show solution
By using the linearity and chain rules we obtain for the first term
âu uHxL = sinHxL,
= cosHxL
âx âf
f HuL = ln u,
=
âu
1 u
for the second term Out[43]=
âu uHxL = cosH4 xL,
= 4 sinH4 xL
âx f HuL =
1
u2 ,
âf =u
âu
2
Finally we get
ây = cotHxL + 4 sinH4 xL cosH4 xL
âx ây = 2 sinH8 xL + cotHxL
or
âx
20.
1 yHxL = 4
tan4 HxL 
1 2
tan2 HxL  lnHcosHxLL
8  Numerical Integration
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By using the linearity and chain rules we obtain Out[47]=
창y 창x
= tanHxL + tan3 HxL sec2 HxL  tanHxL sec2 HxL
창y or
창x
= tan5 HxL