# Example 36 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 36Find 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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.