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

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

Transcript

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
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.