अवकलन के सूत्र तथा सीमाओं का प्रसार ( Formulas of Differentiation )

अवकलन के सूत्र ( Fomulas of Differentiation )Hindi PDF Download

Here Are Some Important Formulas Of Differentiation. These Formulas are very important for those students who are related to Mathematics or physics.So here we provide The PDF File of these very important formulas of Differentiation.

Differentiation formulas
d 𝓍 n / d𝓍
= n 𝓍 n-1
d (Sin𝓍) / d𝓍
= Cos𝓍
d (Cos𝓍) / d𝓍
= - Sin𝓍
d (Tan𝓍) / d𝓍
= Sec 2𝓍
d (Cot𝓍) / d𝓍
= - Cosec 2𝓍
d (Sec𝓍) / d𝓍
= Sec𝓍 ∙ Tan𝓍
d (Cosec𝓍) / d𝓍
= - Cosec𝓍 ∙ Cot𝓍
d (Sin - 1𝓍) / d𝓍
=
1 / 1 - 𝓍 2
d (Cos - 1𝓍) / d𝓍
=
- 1 / 1 - 𝓍 2
d (Tan - 1𝓍) / d𝓍
=
1 / 1 + 𝓍 2
d (Cot - 1𝓍) / d𝓍
=
- 1 / 1 + 𝓍 2
d (Sec - 1𝓍) / d𝓍
=
1 / 𝓍𝓍 2 - 1
d (Cosec - 1𝓍) / d𝓍
=
- 1 / 𝓍𝓍 2 - 1
d e 𝓍 / d𝓍
= e 𝓍
d e - 𝓍 / d𝓍
= - e 𝓍
d log 𝓍 / d𝓍
=
1 / 𝓍
d a 𝓍 / d𝓍
= a 𝓍 log𝓍
d 𝓍 / d𝓍
=
1 / 2 𝓍

सीमाओं का प्रसार

e𝓍 = 1 + 𝓍 +
𝓍 2 / 2!
+
𝓍 3 / 3!
+ ....
e-𝓍 = 1 - x +
𝓍 2 / 2!
-
𝓍 3 / 3!
+ ....
a𝓍 = 1 + (𝓍logea ) +
(𝓍logea )2 / 2!
+
(𝓍logea )3 / 3!
+ ....
loge(1 + 𝓍 ) = 𝓍 -
𝓍 2 / 2
+
𝓍3 / 3
- ....
loge(1 - 𝓍 ) = - 𝓍 -
𝓍 2 / 2
-
𝓍3 / 3
- ....
Sin 𝓍 = 𝓍 -
𝓍 3 / 3!
+
𝓍5 / 5!
- ....
Cos 𝓍 = 1 -
𝓍 2 / 2!
+
𝓍4 / 4!
- ....
tan 𝓍 = 𝓍 +
𝓍 3 / 3
+
2𝓍5 / 15
+ ....
(1 ± 𝓍 )n = 1 ± n𝓍 +
n(n-1) / 2!
𝓍2 ± ....
( 1 +
1 / 𝓍
)𝓍 = e ( 1 +
𝓍 / 2
+
11 / 24
𝓍2 + ...... )
Σn = 1 + 2 + 3 + .... + n = n
(n+1) / 2
Σn2 = 12 + 22 + 32 + .... + n2 = n
(n+1)(2n+1) / 6
Σn3 = 13 + 23 + 33 + .... + n3 = n
{n (n+1)}2 / 4

सीमाओं सम्बंधित कुछ महत्वपूर्ण परिणाम

lim𝓍 → 0
Sin 𝓍 / 𝓍
= 1
lim𝓍 → 0
Cos 𝓍 = 1
lim𝓍 → 0
tan 𝓍 / 𝓍
= 1
lim𝓍 → 0
e𝓍 - 1 / 𝓍
= 1
lim𝓍 → 0
b𝓍 - 1 / 𝓍
= logeb(b≠0)
lim𝓍 → 0
loge [1+𝓍] / 𝓍
= 1
lim𝓍 → 0
[1 + 𝓍]1/𝓍 = e
lim𝓍 → ∞
[ 1 +
1 / 𝓍
]𝓍 = e
lim𝓍 → a
𝓍m - am / 𝓍 - a
= mam - 1
lim𝓍 → a
𝓍m - am / 𝓍n - an
=
m / n
am - n
lim𝓍 → 0
Sin-1𝓍 = 𝓍 +
𝓍3 / 3!
+
9𝓍5 / 5!
+ .....
lim𝓍 → 0
tan-1𝓍 = 𝓍 -
𝓍3 / 3
+
𝓍5 / 5
- .....

Tags: avkalan sutra, avkalan ka formula, avkalan formula, avkalan, samakalan avkalan formula


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