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Ex 5.2, 5 - Differentiate sin(ax+b)/cos(cx+d) - Class 12 CBSE

Ex 5.2, 5 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.2, 5 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Ex 5.2, 5 Differentiate the functions with respect to π‘₯ : sin⁑〖 (π‘Žπ‘₯ + 𝑏)γ€—/cos⁑〖 (𝑐π‘₯ + 𝑑)γ€— Let 𝑦 = sin⁑〖 (π‘Žπ‘₯ + 𝑏)γ€—/cos⁑〖 (𝑐π‘₯ + 𝑑)γ€— Let 𝑒 = sin⁑〖 (π‘Žπ‘₯+𝑏)γ€— & 𝑣=cos⁑〖 (𝑐π‘₯+𝑑)γ€— ∴ π’š = 𝒖/𝒗 We need to find derivative of 𝑦 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑒/𝑣)^β€² 𝑑𝑦/𝑑π‘₯ = (𝑒^β€² 𝑣 βˆ’ γ€– 𝑣〗^β€² 𝑒)/𝑣^2 Finding 𝒖’ 𝑒=sin⁑〖 (π‘Žπ‘₯+𝑏)γ€— Derivative of 𝑒 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑒/𝑑π‘₯ =𝑑(sin⁑〖 (π‘Žπ‘₯+𝑏)γ€— )/𝑑π‘₯ γ€–=cos 〗⁑(π‘Žπ‘₯+𝑏) . 𝑑(π‘Žπ‘₯ + 𝑏)/𝑑π‘₯ γ€–=cos 〗⁑(π‘Žπ‘₯+𝑏) (π‘Ž +0) γ€–=𝐚 𝒄𝒐𝒔 〗⁑(𝒂𝒙+𝒃) Finding 𝒗’ 𝑣=cos⁑〖 (𝑐π‘₯+𝑑)γ€— Derivative of 𝑣 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑣/𝑑π‘₯ = (𝑑(cos⁑〖 (𝑐π‘₯ + 𝑑)γ€— )^β€² )/𝑑π‘₯ γ€–=βˆ’si𝑛 〗⁑(𝑐π‘₯+𝑑) . 𝑑(𝑐π‘₯ + 𝑑)/𝑑π‘₯ γ€–=βˆ’sin 〗⁑(𝑐π‘₯+𝑏) (𝑐+0) γ€–=βˆ’πœ π’”π’Šπ’ 〗⁑(𝒄𝒙+𝒃) Now, π’…π’š/𝒅𝒙 = (𝒖^β€² 𝒗 βˆ’ γ€– 𝒗〗^β€² 𝒖)/𝒗^𝟐 = (γ€–a cos 〗⁑〖(π‘Žπ‘₯ + 𝑏) .γ€– cos〗⁑〖 (𝑐π‘₯ + 𝑑)γ€— βˆ’ (βˆ’π‘ γ€–sin 〗⁑〖(𝑐π‘₯ + 𝑑) γ€— ) γ€— (γ€–sin 〗⁑〖(π‘Žπ‘₯ + 𝑏)γ€— ) )/(cos⁑〖 (𝑐π‘₯ + 𝑑)γ€— )^2 = (γ€–a cos 〗⁑〖(π‘Žπ‘₯ + 𝑏) .γ€– cos〗⁑(𝑐π‘₯ + 𝑑) + 𝑐 . γ€–sin 〗⁑(𝑐π‘₯ + 𝑑) γ€— sin⁑〖(π‘Žπ‘₯ + 𝑏)γ€— )/cos^2⁑(𝑐π‘₯ + 𝑑) = (γ€–a cos 〗⁑〖(π‘Žπ‘₯ + 𝑏) .γ€– cos〗⁑〖 (𝑐π‘₯ + 𝑑)γ€— γ€— )/cos^2⁑(𝑐π‘₯ + 𝑑) + (𝑐 . γ€–sin 〗⁑〖(𝑐π‘₯ + 𝑑) γ€—. sin⁑〖 (π‘Žπ‘₯ + 𝑏)γ€— )/cos^2⁑(𝑐π‘₯ + 𝑑) = a cos (π‘Žπ‘₯+ 𝑏) 𝟏/𝒄𝒐𝒔⁑〖 (𝒄𝒙 + 𝒅)γ€— + 𝑐 . γ€–sin 〗⁑(π‘Žπ‘₯+𝑏) . (γ€–π’”π’Šπ’ 〗⁑〖(𝒄𝒙 + 𝒅)γ€— )/𝒄𝒐𝒔⁑〖 (𝒄𝒙 + 𝒅)γ€— 𝟏/𝒄𝒐𝒔⁑〖 (𝒄𝒙 + 𝒅)γ€— = 𝒂 𝒄𝒐𝒔 (π‘Žπ‘₯+ 𝑏) .𝒔𝒆𝒄⁑〖(𝑐π‘₯+𝑑)γ€— + 𝒄 . γ€–π’”π’Šπ’ 〗⁑(π‘Žπ‘₯+𝑏).〖𝒕𝒂𝒏 〗⁑(𝑐π‘₯+𝑑).𝒔𝒆𝒄⁑(𝑐π‘₯+𝑑)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.