Uploaded by Ishaan Sharma

DIFFERENTIATION - HIGHER ORDER

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DIFFERENTIATION
−๐Ÿ ๐’•
, ๐’š = √๐’‚๐œ๐จ๐ฌ
−๐Ÿ ๐’•
1
If ๐’™ = √๐’‚๐ฌ๐ข๐ง
2
If ๐’™ = ๐œ๐จ๐ฌ ๐’• + ๐’• ๐ฌ๐ข๐ง ๐’• , ๐’š = ๐ฌ๐ข๐ง ๐’• − ๐’• ๐œ๐จ๐ฌ ๐’• , show that
3
If ๐’™ = ๐’‚ ๐œ๐จ๐ฌ ๐’• + ๐’ƒ ๐ฌ๐ข๐ง ๐’• , ๐’š = ๐’‚ ๐ฌ๐ข๐ง ๐’• − ๐’ƒ ๐œ๐จ๐ฌ ๐’• , prove that ๐’š๐Ÿ
4
If ๐’™ = ๐’‚(๐Ÿ − ๐œ๐จ๐ฌ ๐œฝ) , ๐’š = ๐’‚(๐œฝ + ๐ฌ๐ข๐ง ๐œฝ) ๐’‘๐’“๐’๐’—๐’† ๐’•๐’‰๐’‚๐’•
5
If ๐’™ = ๐ฌ๐ข๐ง−๐Ÿ (๐Ÿ+๐’•๐Ÿ ) , ๐’š = ๐ญ๐š๐ง−๐Ÿ (๐Ÿ−๐’•๐Ÿ ) , ๐’• > 1 , ๐‘๐‘Ÿ๐‘œ๐‘ฃ๐‘’ ๐‘กโ„Ž๐‘Ž๐‘ก
6
If ๐’š = ๐‘จ๐’†๐’Ž๐’™ + ๐‘ฉ๐’†๐’๐’™ ,prove that
7
If ๐’š = ๐’†๐’‚ ๐œ๐จ๐ฌ
8
If
9
๐’‚ > 0 , −1 < ๐‘ก < 1 show that
๐Ÿ๐’•
๐’™๐Ÿ
๐’‚๐Ÿ
−๐Ÿ ๐’™
๐’…๐Ÿ ๐’š
๐’…๐’™๐Ÿ
=
๐’…๐Ÿ ๐’š
− (๐’Ž + ๐’)
๐’…๐’™๐Ÿ
๐’…๐’š
๐’…๐’™
=
−๐’š
๐’™
๐’”๐’†๐’„๐Ÿ‘ ๐’•
๐’•
๐’…๐Ÿ ๐’š
๐’…๐’™๐Ÿ
−๐’™
−๐Ÿ
๐’…๐’š
๐’…๐’™
+๐’š = ๐ŸŽ
๐œฝ
๐œฝ
= ๐Ÿ’๐’‚ ๐ฌ๐ž๐œ (๐Ÿ) ๐’„๐’๐’”๐’†๐’„๐Ÿ‘ (๐Ÿ)
๐Ÿ๐’•
๐’…๐’š
๐’…๐’™
= −๐Ÿ
+ ๐’Ž๐’๐’š = ๐ŸŽ
๐’‘๐’“๐’๐’—๐’† ๐’•๐’‰๐’‚๐’• (๐Ÿ − ๐’™๐Ÿ )๐’š๐Ÿ − ๐’™๐’š๐Ÿ − ๐’‚๐Ÿ ๐’š = ๐ŸŽ
๐’š๐Ÿ
๐’…๐Ÿ ๐’š
๐’…๐’™๐Ÿ
+ ๐’ƒ๐Ÿ = ๐Ÿ , prove that
−๐’ƒ๐Ÿ’
= ๐’‚๐Ÿ ๐’š๐Ÿ‘
If y = ๐ญ๐š๐ง ๐’™ + ๐ฌ๐ž๐œ ๐’™ , prove that (๐Ÿ − ๐ฌ๐ข๐ง ๐’™)๐Ÿ
๐Ÿ
10
If y = [๐’๐’๐’ˆ (๐’™ + √๐’™๐Ÿ + ๐Ÿ)]
11
Prove that ๐’™๐Ÿ‘ ๐’š๐Ÿ = (๐’™ ๐’š๐Ÿ − ๐’š)๐Ÿ , if ( a+ bx ) ๐’†
๐’…๐Ÿ ๐’š
๐’…๐’™๐Ÿ
= ๐œ๐จ๐ฌ ๐’™ .
then prove that , ( 1 + ๐’™๐Ÿ )๐’š๐Ÿ + ๐’™ ๐’š๐Ÿ = ๐Ÿ
๐’š⁄
๐’™
=๐’™
๐’…๐Ÿ ๐’š
12 y = ๐’†๐’™ ๐ญ๐š๐ง−๐Ÿ ๐’™ , prove that ( 1+๐’™๐Ÿ )๐’…๐’™๐Ÿ − ๐Ÿ(๐Ÿ − ๐’™ + ๐’™๐Ÿ )
๐’…๐Ÿ ๐’š
๐’…๐Ÿ ๐’š
๐’…๐’™๐Ÿ
๐’…๐’š
๐’…๐’™
๐Ÿ
(๐Ÿ − ๐’™)๐Ÿ ๐’š = ๐ŸŽ
๐’•
13
Find
14
If ๐’š๐Ÿ‘ + ๐’™๐Ÿ‘ − ๐Ÿ‘๐’‚๐’™๐’š = ๐ŸŽ , prove that
15
If ๐’™ = ๐ฌ๐ž๐œ ๐œฝ − ๐œ๐จ๐ฌ ๐œฝ , ๐’š = ๐’”๐’†๐’„๐’ ๐œฝ − ๐’„๐’๐’”๐’ ๐œฝ ,prove that (๐’™๐Ÿ + ๐Ÿ’)(๐’š| ) = ๐’๐Ÿ (๐’š๐Ÿ + ๐Ÿ’)
๐’…๐’™๐Ÿ
, given that , x = a [๐œ๐จ๐ฌ ๐’• +
๐’…๐’š
+
๐’…๐’™
๐Ÿ
๐’…๐Ÿ ๐’š
๐’…๐’™๐Ÿ
๐’๐’๐’ˆ ๐’•๐’‚๐’๐Ÿ ( )] and y = a ๐ฌ๐ข๐ง ๐’•
๐Ÿ
−๐Ÿ๐’‚๐Ÿ‘ ๐’™๐’š
= (๐’š๐Ÿ −๐’‚๐’™)๐Ÿ‘
๐Ÿ
16 If ๐’™ = ๐Ÿ ๐œ๐จ๐ฌ ๐œฝ − ๐œ๐จ๐ฌ ๐Ÿ๐œฝ , ๐’š = ๐Ÿ ๐ฌ๐ข๐ง ๐œฝ − ๐ฌ๐ข๐ง ๐Ÿ๐œฝ ๐’•๐’‰๐’†๐’
17
If ๐’š = [√๐’™ + ๐Ÿ + √๐’™ − ๐Ÿ]
๐’Ž
๐’…๐’š
๐’…๐’™
= ๐ญ๐š๐ง(๐’‘๐œฝ) , ๐’‡๐’Š๐’๐’… ๐’‘
๐Ÿ
, prove that (๐’™๐Ÿ − ๐Ÿ)๐’š๐Ÿ + ๐’™๐’š๐Ÿ = ๐Ÿ’ ๐’Ž๐Ÿ ๐’š
18
If ๐’‡(๐’™) = |๐’™|๐Ÿ‘ show that ๐’‡|| (๐’™) exist for all real , then find ๐’‡|| (๐’™) .
19
If ๐’š = ๐’™ +
๐Ÿ
๐’™+
๐’™+
๐Ÿ
prove that
๐Ÿ
๐Ÿ
๐’™+๐’™ ++++++++∞
20
Prove that the derivative of ๐ญ๐š๐ง−๐Ÿ (
21∗
If ๐’š = ๐ฌ๐ข๐ง−๐Ÿ
๐Ÿ
+ ๐ญ๐š๐ง−๐Ÿ
๐Ÿ
√๐Ÿ+๐’™
√๐Ÿ+๐’™๐Ÿ −๐Ÿ
√๐Ÿ+๐’™๐Ÿ −๐Ÿ
๐’™
๐’…๐’š
๐’…๐’™
๐’™
๐’š
= ๐Ÿ๐’š−๐’™
) ๐’˜๐’Š๐’•๐’‰ ๐’“๐’†๐’”๐’‘๐’†๐’„๐’• ๐’•๐’ ๐ญ๐š๐ง−๐Ÿ (
, then show that
๐’…๐’š
๐’…๐’™
−๐Ÿ
= ๐Ÿ(๐Ÿ+๐’™๐Ÿ )
๐Ÿ๐’™√๐Ÿ−๐’™๐Ÿ
)
๐Ÿ−๐Ÿ๐’™๐Ÿ
๐Ÿ
๐’‚๐’• ๐’™ = ๐ŸŽ ๐’Š๐’” ๐Ÿ’
๐’…๐’š
22
If ๐’™ = ๐’‚ ๐ฌ๐ข๐ง ๐Ÿ๐œฝ(๐Ÿ + ๐œ๐จ๐ฌ ๐Ÿ๐œฝ) ๐’‚๐’๐’… ๐’š = ๐’ƒ ๐œ๐จ๐ฌ ๐Ÿ๐œฝ(๐Ÿ − ๐œ๐จ๐ฌ ๐Ÿ๐œฝ) , find ๐’…๐’™
23
Find
24
If ๐’™ =
25
If ๐’™ = ๐’†
26∗
if ๐’š = ๐ฌ๐ข๐ง−๐Ÿ (๐Ÿ+(๐Ÿ‘๐Ÿ”)๐’™ ) prove that
27
If ๐’š = ๐’๐’๐’ˆ(๐Ÿ + ๐œ๐จ๐ฌ ๐’™) ,prove that
28
If ๐’š = (๐ฌ๐ข๐ง−๐Ÿ ๐’™) + (๐œ๐จ๐ฌ−๐Ÿ ๐’™) , prove that (๐Ÿ − ๐’™๐Ÿ ) ๐’…๐’™๐Ÿ − ๐’™ ๐’…๐’™ = ๐Ÿ’
29∗
If ๐’š =
30
If ๐’™ = ๐’†๐’š , prove that
31
Find the derivative of :
32
If ๐’š =
33
If ๐’™ = ๐’‚(๐œ๐จ๐ฌ ๐Ÿ๐’• + ๐Ÿ๐’• ๐ฌ๐ข๐ง ๐Ÿ๐’•) ๐’‚๐’๐’… ๐’š = ๐’‚(๐ฌ๐ข๐ง ๐Ÿ๐’• − ๐Ÿ๐’• ๐œ๐จ๐ฌ ๐Ÿ๐’•) , then find ๐’…๐’™๐Ÿ .
34
If ๐’™ = ๐’‚ ๐ฌ๐ข๐ง ๐’‘๐’• , ๐’š = ๐’ƒ ๐œ๐จ๐ฌ ๐’‘๐’• , show that (๐’‚๐Ÿ − ๐’™๐Ÿ )๐’š
๐’…๐’š
๐’…๐’™
๐Ÿ ๐’‚
๐Ÿ
, ๐’š = ๐’‚(๐’•+ ๐’• ) , ๐’‚ > 0
, when ๐’™ = (๐’• + ๐’• )
๐Ÿ+๐’๐’๐’ˆ๐’•
๐’•๐Ÿ
๐ญ๐š๐ง−๐Ÿ (
๐Ÿ‘+๐Ÿ๐’๐’๐’ˆ๐’•
๐’•
๐’š=
๐’š−๐’™๐Ÿ
)
๐’™๐Ÿ
, prove that
๐’…๐’š
๐’…๐’™
๐Ÿ๐’™+๐Ÿ ๐Ÿ‘๐’™
๐Ÿ
๐Ÿ
√๐Ÿ“
๐ญ๐š๐ง−๐Ÿ (
๐Ÿ
๐’™
๐’…๐’š
๐’…๐’™
=
๐Ÿ .๐Ÿ”๐’™ ๐’๐’๐’ˆ๐Ÿ”
๐Ÿ+(๐Ÿ‘๐Ÿ”)๐’™
๐’…๐Ÿ‘ ๐’š
๐’…๐Ÿ ๐’š
+
๐Ÿ‘
๐’…๐’™
๐’…๐’™๐Ÿ
๐’™
, prove that
๐’…๐’š
๐’…๐’™
๐’…๐’š
. ๐’…๐’™ = ๐ŸŽ
๐’…๐Ÿ ๐’š
๐ญ๐š๐ง ๐Ÿ) , ๐’”๐’‰๐’๐’˜ ๐’•๐’‰๐’‚๐’•
๐’™
๐Ÿ
= ๐’™[๐’”๐’†๐’„๐Ÿ (๐’๐’๐’ˆ๐’™) + ๐Ÿ + ๐Ÿ ๐ญ๐š๐ง(๐’๐’๐’ˆ๐’™)]
๐Ÿ
√๐Ÿ“
๐’…๐’š ๐Ÿ
๐’…๐’š
, ๐’• > 0 then show that , ๐’š ๐’…๐’™ = ๐Ÿ๐’™ (๐’…๐’™) + ๐Ÿ
๐’…๐’š
๐’…๐’™
=
๐’…๐’š
๐Ÿ
๐Ÿ‘+๐Ÿ ๐œ๐จ๐ฌ ๐’™
๐’™−๐’š
= ๐’™๐’๐’๐’ˆ๐’™
๐ฌ๐ข๐ง(๐’™๐’™ ) + (๐ฌ๐ข๐ง ๐’™)๐’™
๐’…๐’š
√๐Ÿ+๐’š๐Ÿ’
+
๐’…๐’™
√๐Ÿ+๐’™๐Ÿ’
=๐ŸŽ
๐’…๐Ÿ ๐’š
๐’…๐Ÿ ๐’š
๐’…๐’™๐Ÿ
+ ๐’ƒ๐Ÿ = ๐ŸŽ .
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