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All of the need formulas for Differential calculus and integration
Typology: Exams
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= lim
∆𝑥→ 0
𝑤ℎ𝑒𝑟𝑒: 𝑐 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡; 𝑢, 𝑣 𝑎𝑛𝑑 𝑤 𝑎𝑟𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛s
𝑛
𝑛− 1
2
(sin 𝑢) = cos 𝑢 𝑑𝑢
(cos 𝑢) = − sin 𝑢 𝑑𝑢
(tan 𝑢) = 𝑠𝑒𝑐
2
(cot 𝑢) = −𝑐𝑠𝑐
2
(sec 𝑢) = sec 𝑢 tan 𝑢 𝑑𝑢
(csc 𝑢) = −csc 𝑢 cot 𝑢 𝑑𝑢
− 1
2
− 1
2
− 1
2
− 1
2
− 1
2
− 1
2
(sinh 𝑢) = cosh 𝑢 𝑑𝑢
(cosh 𝑢) = sinh 𝑢 𝑑𝑢
(tanh 𝑢) = 𝑠𝑒𝑐ℎ
2
(coth 𝑢) = −𝑐𝑠𝑐ℎ
2
(sech 𝑢) = −sech 𝑢 tanh 𝑢 𝑑𝑢
(csch 𝑢) = −csch 𝑢 coth 𝑢 𝑑𝑢
− 1
2
− 1
2
− 1
2
− 1
2
− 1
2
− 1
2
𝑢
𝑢
ln 𝑎 𝑑𝑢
𝑢
𝑢
(log
𝑎
log 𝑒 𝑑𝑢 =
2 ln 𝑎
𝑢 ln 𝑎
(ln 𝑢) =
lim
𝑥→ 0
= lim
𝑥→ 0
𝑢 ± lim
𝑥→ 0
𝑣 ± lim
𝑥→ 0
lim
𝑥→ 0
(𝑢𝑣𝑤) = (lim
𝑥→ 0
𝑢) (lim
𝑥→ 0
𝑣) (lim
𝑥→ 0
lim
𝑥→ 0
lim
𝑥→ 0
lim
𝑥→ 0
lim
𝑥→ 0
= 𝑐 lim
𝑥→ 0
lim
𝑥→ 0
𝑛
lim
𝑥→ 0
𝑛
lim
𝑥→∞
𝑋
= ∞ lim
𝑥→−∞
𝑋
lim
𝑥→∞
ln 𝑥 = ∞ lim
𝑥→∞
𝑐
𝑥
𝑛
lim
𝑥→∞
𝑥
√𝑥!
𝑥
= 𝑒 lim
𝑥→∞
𝑘
𝑥
)
𝑥
𝑘
lim
𝑥→∞
1
𝑥
)
𝑥
1
𝑒
lim
𝑥→∞
√
2 𝜋𝑥
𝑥!
1
𝑥
lim
𝑥→∞
𝑥!
𝑥
𝑥
𝑒
−𝑥
√
𝑥
2 𝜋 lim
𝑥→∞
log
𝑎
1
𝑥
𝑥
log
𝑎
lim
𝑥→ 0
log 𝑒
( 1 +𝑥)
𝑥
= 1 lim
𝑥→ 0
𝑥
log
𝑎
( 1 +𝑥)
1
log
𝑎
𝑒
lim
𝑥→ 0
𝑎
𝑥
− 1
𝑥
= ln 𝑎 (𝑎 > 0 ) lim
𝑥→ 0
sin 𝑥
𝑥
lim
𝑥→ 0
tan 𝑥
𝑥
= 1 lim
𝑥→ 0
1 −cos 𝑥
𝑥
lim
𝑥→ 0
1 −cos 𝑥
𝑥
2
1
2
lim
𝑥→ 0
sin
− 1
𝑥
𝑥
lim
𝑥→ 0
tan
− 1
𝑥
𝑥
= 1 lim
𝑥→ 1
(cos
− 1
𝑥)
2
1 −𝑥
𝐿𝑖𝑚𝑖𝑡𝑠 𝑎𝑡 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑎𝑠 lim
𝑥→∞
lim
𝑥→∞
lim
𝑥→∞
2
3
sin 0
cos 1
tan 0
cot ∞ √ 3
sec 1
csc ∞ 2 √ 2
arcsin 0 90
arccos 90 0
arctan 0 45
arccot ∞ 0.
arcsec 1 0.
arccsc ∞ 0.
ln 𝑥 + ln 𝑦 = ln(𝑥𝑦) ln 𝑥 − ln 𝑦 = ln(
𝑥
𝑦
ln 𝑥
𝑚
= 𝑚 ln 𝑥 ln 𝑒
𝑘
ln 𝑥
= 𝑥 ln 1 = 0 ln 𝑒 = 1
sinh 𝑥 =
𝑒
𝑥
−𝑒
−𝑥
2
cosh 𝑥 =
𝑒
𝑥
+𝑒
−𝑥
2
2
2
𝑥 = 1 cosh 𝑥 + sinh 𝑥 = 𝑒
𝑥
2
2
𝑥 cosh 𝑥 − sinh 𝑥 = 𝑒
−𝑥
2
2
cosh(𝑥 ± 𝑦) = cosh 𝑥 cosh 𝑦 ± sinh 𝑥 sinh 𝑦
sinh
= sinh 𝑥 cosh 𝑦 ± sinh 𝑦 cosh 𝑥
sinh 2 𝑥 = 2 sinh 𝑥 cosh 𝑥
cosh 2 𝑥 = 𝑐𝑜𝑠ℎ
2
2
sinh
− 1
𝑥 = ln(𝑥 + √
2
cosh
− 1
𝑥 = ln(𝑥 +
2
tanh
− 1
1
2
ln(
1 +𝑥
1 −𝑥
)
coth
− 1
1
2
ln(
𝑥+ 1
𝑥− 1
)
𝑛
𝑛+ 1
∫ sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶
∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶
∫ tan 𝑥 𝑑𝑥 = − ln|cos 𝑥| + 𝐶 = ln|sec 𝑥| + 𝐶
∫ cot 𝑥 𝑑𝑥 = ln
sin 𝑥
∫ sec 𝑥 𝑑𝑥 = ln
sec 𝑥 + tan 𝑥
∫ csc 𝑥 𝑑𝑥 = ln
csc 𝑥 − cot 𝑥
∫ sec
2
𝑥 𝑑𝑥 = tan 𝑥 + 𝐶
∫ csc
2
𝑥 𝑑𝑥 = − cot 𝑥 + 𝐶
∫ sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝐶
csc 𝑥 cot 𝑥 = − csc 𝑥 + 𝐶
2
2
𝑑𝑢 = sin
− 1
2
2
tan
− 1
2
2
sec
− 1
− 1
𝑑𝑢 = ln
𝑢
𝑢
𝑢
𝑢
ln 𝑎
∫ √𝑢
2
± 𝑎
2
𝑑𝑢 =
1
2
[𝑢√𝑢
2
± 𝑎
2
± 𝑎
2
ln |𝑢 +
√ 𝑢
2
√𝑢
2
± 𝑎
2
|]
2
2
𝑑𝑢 = ln
2
2
2
2
2
2
2
sin
− 1
2
2
ln |
2
2
ln |
𝐴
𝐵
𝑅
𝐿
𝑇
𝐴
𝐵
𝑏
𝑎
𝑇
𝐴
2
𝐵
2
𝑏
𝑎
𝑇
𝑅
2
𝐿
2
𝑏
𝑎
𝑇
𝑅
𝐿
𝑏
𝑎
2
𝑏
𝑎
2
2
1
2
𝑏
𝑎
𝑏
𝑎
2
𝑏
𝑎