Example 36 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at May 29, 2018 by Teachoo
Transcript
Example 36 Find ππ¦/ππ₯ , if π₯ = π (π+sinβ‘π), π¦ = π (1 β cosβ‘π) ππ¦/ππ₯ = ππ¦/ππ₯ Γ ππ/ππ ππ¦/ππ₯ = ππ¦/ππ Γ ππ/ππ₯ ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) Therefore ππ¦/ππ₯ = (ππ¦/ππ)/(ππ₯/ππ) ππ¦/ππ₯ = π" " (sinβ‘π )/π" " (1 +γ cosγβ‘π ) ππ¦/ππ₯ = sinβ‘π/(1 +γ cosγβ‘π ) ππ¦/ππ₯ = (γπ πππγβ‘γ π½/πγ .γ πππγβ‘γ π½/πγ)/(1 + γπ γπππγ^π γβ‘γπ½/πγ β π) ππ¦/ππ₯ = (γ2 sinγβ‘γ π/2γ .γ cosγβ‘γ π/2γ)/γ2 cos^2 γβ‘γπ/2γ ππ¦/ππ₯ = (sinβ‘γ π/2γ )/γcos γβ‘γπ/2γ π π/π π = πππ§β‘γπ½/πγ
Examples
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11
Example 12
Example 13
Example 14
Example 15
Example 16
Example 17
Example 18
Example 19
Example 20
Example 21
Example 22
Example 23 Important
Example 24
Example 25
Example 26 Important
Example 27
Example 28
Example 29
Example 30 Important
Example 31
Example 32 Important
Example 33 Important
Example 34
Example 35
Example 36 You are here
Example 37
Example 38
Example 39
Example 40 Important
Example 41 Important
Example 42 Important
Example 43
Example 44 Important
Example 45 Important
Example 46
Example 47 Important
Example 48
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.