Calculus

 Last Name 1
Name
Professor
Subject
Date
Calculus
Question No. 1
Given ???? = ???????????? (????????????
−????(????
????????))find ????
′
???????????????? ???????? ????????
????????
. And simplify your answer.
Solution
???? = ????????????(????????????
−1(????
4????))
Using chain rule
????????
???????? =
????????
???????? .
????????
???????? … . . … . . .. (1)
Consider ???? = (????????????
−1(????
4????)) ???????????? ???? = ????????????(????)
????????
???????? =
????
???????? (????????????(????)) = sec2(????) … … .. (2)
Substituting back the value of “u” in equation 2 gives
????????
???????? = sec2 (( ???????????? ???????????? (????
4????))) … …… (3)
????????
???????? =
????
???????? ((???????????? ???????????? (????
4????)))
Using product rule in equation 1
=
????
????????
(arccos(????
4????))
????
???????? (????
4????
)
As we know that ????
????????
arccos(????) = (−
1
√1−(????)
2
)
 Last Name 2
????????
???????? = (−
1
√1 − (????
4????)
2
) × 4(????
4????) = −
4????
4????
√1 − (????
4????)
2
…… (4)
Substituting equation 3 and 4 in equation 1 gives
????
′ =
????????
???????? = sec2 (( ???????????? ???????????? (????
4????))) (−
4????
4????
√1 − (????
4????)
2
) … … . . (5)
Simplification:
Using the mathematical rule sec (( ???????????? ???????????? (????))) =
1
????
in equation 5 gives
sec2 (( ???????????? ???????????? (????
4????))) = (
1
????
4????
)
2
Simplified form becomes
????
′ = (
1
????
4????
)
2
(−
4????
4????
√1 − (????
4????)
2
)
????
′ = −
4
????
4????√1 − (????
8????)
Question No. 2
Evaluate the following
∫
????????????????
???????????????? + ???????????? ????
????
????
????
???????? =
????
????
Solution
Let
???? = ∫
???????????? ????
???????????????? + ???????????? ????
????
2
0
???????? … … … …(1)
Multiplying dominator and denominator with factor (sin ???? − cos ????) in equation 1.
 Last Name 3
???? = ∫
???????????? ???? (sin ???? − cos ????)
(???????????????? + cos ????)(sin ???? − cos ????)
????
2
0
????????
= ∫
???????????? ???? (sin ???? − cos ????)
(sin2 ???? − cos2 ????)
????
2
0
????????
As we know that
(sin2 ???? − cos2 ????) = cos (2????)
It becomes
(cos2 ???? − sin2 ????) = − cos(2????) … … … … (2)
???? = ∫
???????????? ???? (sin ???? − cos ????)
(sin2 ???? − cos2 ????)
????
2
0
???????? = ∫
− ???????????? ???? cos ???? + sin2 ????
(sin2 ???? − cos2 ????)
????
2
0
???????? … … … (3)
Inserting equation 2 in equation 3
???? = ∫ [
− ???????????? ???? cos ????
− cos(2????)
+
sin2 ????
cos(2????)
]
????
2
0
????????
= ∫ [
???????????? ???? cos ????
cos (2????)
+
sin2 ????
cos (2????)
]
????
2
0
???????? … … …… (4)
???? = ????1 + ????2
Inserting equation 5 in equation 4
???????????? ???? cos ???? =
sin (2????)
2
… … … … (5)
????1 = ∫ [
???????????? ???? cos ????
cos (2????)
]
????
2
0
???????? = ∫ [
sin (2????)
2
cos (2????)
]
????
2
0
????????
 Last Name 4
????1 =
1
2
∫ [
sin (2????)
cos (2????)
]
????
2
0
????????
???????????????????????????????????????????????? ???? = cos(2????)
????1 =
1
2
∫ −
1
2????
???????? = −
1
4
∫
1
????
???????? = −
1
4
ln|????| … … … (6)
Substituting back in equation 6
????1 = −
1
4
ln|cos(2????)| … … … (6)
????2 = ∫ [
sin2 ????
cos (2????)
]
????
2
0
???????? … … … … (7)
Using the following identity
sin2 ???? =
1 − 2 cos (2????)
2
Equation 7 becomes
????2 = ∫ [
1− 2 cos (2????)
2
cos(2????)
]
????
2
0
????????
????2 =
1
2
∫ [
1 − 2 cos (2????)
cos (2????)
]
????
2
0
???????? =
1
2
∫ [
1
cos (2????)
−
cos (2????)
cos (2????)
]
????
2
0
????????
????2 =
1
2
∫ [
1
cos (2????)
− 1]
????
2
0
????????
????2 =
1
2
[∫ [
1
cos(2????)
] ???????? − ∫[1] ????????]
 Last Name 5
????2 =
1
2
[(
1
2
ln|tan(2????)| + sec(2????)) − ????]… … . … (8)
Using equation 6 and equation 8
???? = −
1
4
ln|cos(2????)| +
1
2
[(
1
2
ln|tan(2????)| + sec(2????)) − ????]
Computing the boundaries
lim ???? → 0
+
(−
1
4
ln|cos(2????)| +
1
2
[(
1
2
ln|tan(2????)| + sec(2????)) − ????]) = 0
lim
???? →
????
2
− (−
1
4
ln|cos(2????)| +
1
2
[(
1
2
ln|tan(2????)| + sec(2????)) − ????]) =
????
4
???? =
????
4
− 0 =
????
4
Hence proved
∫
???????????? ????
???????????????? + ???????????? ????
????
2
0
???????? =
????
4
Question No. 3
Evaluate the following limit. Use limit theorem not ???? − ???? techniques. If any of them fail to
exist, say so and say why.
???????????????? → ????
+
√???? − ????
????
???? − ????
Solution
First simplifying the equation by factoring ????
2 − 1 = (???? + 1)(???? − 1) and inserting in the
equation.
√???? − 1
????
2 − 1
=
√???? − 1
(???? + 1)(???? − 1)
=
1
(???? + 1)√???? − 1
Using x approaching to 1 from the right side the condition becomes 
 Last Name 6
???? > 1 → (???? + 1)√ 


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