Example 36 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 13, 2021 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) Calculating π π/π π½ π¦ = π (1 β cosβ‘π) ππ¦/ππ = π(π (1 β cosβ‘π))/ππ ππ¦/ππ = π (0 β(βsinβ‘π )) ππ¦/ππ = π (πππβ‘π½ ) Calculating π π/π π½ π₯ = π (π+π ππβ‘π) ππ₯/ππ = π(π(π + π ππβ‘π))/ππ ππ₯/ππ = π (ππ/ππ+π(sinβ‘π )/(ππ )) ππ₯/ππ = π (π+ππ¨π¬β‘π½ ) 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 Rough We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by π/2 sin ΞΈ = 2 πππβ‘γπ½/πγ πππβ‘γπ½/πγ and cos 2ΞΈ = 2cos2 ΞΈ β 1 Replacing ΞΈ by π/2 cos ΞΈ = 2cos2 π½/π β 1
Examples
Example 1
Example 2
Example 3
Example 4 Important
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 Important
Example 18
Example 19
Example 20 Important
Example 21
Example 22
Example 23
Example 24
Example 25
Example 26 Important
Example 27
Example 28
Example 29 Important
Example 30 Important
Example 31
Example 32 Important
Example 33 Important
Example 34
Example 35
Example 36 Important You are here
Example 37 Important
Example 38
Example 39
Example 40
Example 41
Example 42 Important Deleted for CBSE Board 2021 Exams only
Example 43 Deleted for CBSE Board 2021 Exams only
Example 44 Important
Example 45 Important
Example 46
Example 47 Important
Example 48
Chapter 5 Class 12 Continuity and Differentiability
Serial order wise
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.