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

Transcript

Ex 5.2, 4 (Introduction) Differentiate the functions with respect to x sec (tan ( √π‘₯ )) Finding derivative of sec x Let 𝑦=sec⁑π‘₯ 𝑦 = 1/cos⁑π‘₯ Let 𝑒 = 1 & 𝑣 = cos⁑π‘₯ So, y = 𝑒/𝑣 (y)’ = (𝑒/𝑣)^β€² 𝑦′ = (𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 Putting values 𝑦′ = (γ€–(1)γ€—^β€² cos⁑π‘₯ βˆ’ γ€–(cos⁑〖π‘₯ γ€—)γ€—^β€² 1)/cos^2⁑π‘₯ y’ = (0 γ€–(cos〗⁑〖π‘₯)γ€— βˆ’ (γ€–βˆ’sin〗⁑〖π‘₯) 1γ€—)/(γ€–π‘π‘œπ‘ γ€—^2 π‘₯) = 0 + sin⁑π‘₯/(γ€–π‘π‘œπ‘ γ€—^2 π‘₯) = sin⁑π‘₯/cos⁑π‘₯ . 1/cos⁑π‘₯ = tan⁑π‘₯.sec⁑π‘₯ Hence, (𝒔𝒆𝒄⁑𝒙 )’ =𝒔𝒆𝒄⁑𝒙 𝒕𝒂𝒏⁑𝒙 Ex 5.2, 4 Differentiate the functions with respect to x sec (tan ( √π‘₯ )) Let 𝑦 = sec (tan √π‘₯ ) We need to find Derivative of 𝑦 i.e. 𝑦’ = (sec⁑〖〖(tanγ€—β‘βˆšπ‘₯)γ€— )^β€² = γ€–sec 〗⁑〖(tan⁑√π‘₯)γ€— γ€–tan 〗⁑〖(tan⁑√π‘₯)γ€— (tan⁑√π‘₯ )^β€² = γ€–sec 〗⁑〖(tan⁑√π‘₯)γ€— γ€–tan 〗⁑〖(tan⁑√π‘₯)γ€—. ("sec2 " √π‘₯ " . " (√π‘₯)^β€²) = γ€–sec 〗⁑〖(tan⁑√π‘₯)γ€— γ€–tan 〗⁑〖(tan⁑√π‘₯)γ€—. sec2 " " √π‘₯ Γ— 1/(2√π‘₯) = (sec⁑〖(tan⁑√π‘₯ γ€—)sec⁑〖(tan⁑√π‘₯ γ€—)sec^2⁑√π‘₯ )/(2√π‘₯) Hence π’…π’š/𝒅𝒙 = (𝒔𝒆𝒄⁑〖(π’•π’‚π’β‘βˆšπ’™ γ€—)𝒔𝒆𝒄⁑〖(π’•π’‚π’β‘βˆšπ’™ γ€—)〖𝒔𝒆𝒄〗^πŸβ‘βˆšπ’™ )/(πŸβˆšπ’™)

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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.