Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.2, 7 Differentiate the functions with respect to π‘₯ 2√(cot ( π‘₯2 )) Let 𝑦 = " " 2√(cot ( π‘₯2 )) We need to find derivative of 𝑦 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑦′ = (2√(cot ( π‘₯^2 ) ) )^β€² = 2(√(cot ( π‘₯^2 ) ) )^β€² = 2 1/(2√(cot ( π‘₯^2 ) )) . (cot ( π‘₯^2 ))^β€² = 1/√(cot ( π‘₯^2 ) ) .(βˆ’π‘π‘œπ‘ π‘’π‘^2 (π‘₯^2 )) . ( π‘₯^2 ) = 1/√(cot ( π‘₯^2 ) ) . (βˆ’π‘π‘œπ‘ π‘’π‘^2 (π‘₯^2)) . 2π‘₯ = (βˆ’2π‘₯" " π‘π‘œπ‘ π‘’π‘^2 π‘₯^2)/√(cot ( π‘₯^2 ) ) = (βˆ’2π‘₯)/(sin^2⁑(π‘₯^2 ) . √( γ€–cos 〗⁑〖(π‘₯^2)γ€—/γ€–sin 〗⁑〖(π‘₯^2)γ€— ) ) "= " (βˆ’2π‘₯)/(〖𝑠𝑖𝑛 〗⁑(π‘₯^2 ) Γ— 〖𝑠𝑖𝑛 〗⁑(π‘₯^2 ) √( γ€–cos 〗⁑〖(π‘₯^2)γ€—/γ€–sin 〗⁑〖(π‘₯^2)γ€— ) ) " " = (βˆ’ 2π‘₯ )/(γ€–sin 〗⁑〖(π‘₯)^2 γ€— Γ—βˆš(γ€–sin^2 〗⁑〖(π‘₯^2)γ€— Γ— γ€–cos 〗⁑〖(π‘₯^2)γ€—/sin⁑〖 (π‘₯^2)γ€— )) = (βˆ’ 2π‘₯ )/(γ€–sin 〗⁑〖(π‘₯)^2 γ€— Γ—βˆš(γ€–cos 〗⁑〖(π‘₯^2)γ€— sin⁑〖 (π‘₯^2)γ€— )) = (βˆ’ 2π‘₯ )/(γ€–sin 〗⁑〖(π‘₯)^2 γ€— Γ—βˆš(2/2 Γ— γ€–cos 〗⁑〖(π‘₯^2)γ€— sin⁑〖 (π‘₯^2)γ€— )) = (βˆ’ 2√2 π‘₯ )/(γ€–sin 〗⁑〖(π‘₯)^2 γ€— Γ—βˆš(2 γ€–cos 〗⁑〖(π‘₯^2)γ€— sin⁑〖 (π‘₯^2)γ€— )) = (βˆ’ 𝟐√𝟐 𝒙 )/(〖𝐬𝐒𝐧 〗⁑〖(𝒙)^𝟐 γ€— Γ—βˆš(γ€–π’”π’Šπ’ 〗⁑〖(πŸπ’™^𝟐)γ€— )) (As sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ)

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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.