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

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

Transcript

Ex 5.2, 5 - Chapter 5 Class 12 Continuity and Differentiability Differentiate the functions with respect to x sin (ax + b) / cos (cx + d) Let 𝑦 = sin⁑〖 (π‘Žπ‘₯+𝑏)γ€—/cos⁑〖 (𝑐π‘₯+𝑑)γ€— Let 𝑒 = sin⁑〖 (π‘Žπ‘₯+𝑏)γ€— & 𝑣=cos⁑〖 (𝑐π‘₯+𝑑)γ€— 𝑦 = 𝑒/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) . (𝑑(π‘Žπ‘₯)/𝑑π‘₯ + 𝑑(𝑏)/𝑑π‘₯) γ€–=cos 〗⁑(π‘Žπ‘₯+𝑏) . (π‘Ž . 𝑑π‘₯/𝑑π‘₯ + 𝑑(𝑏)/𝑑π‘₯) γ€–=cos 〗⁑(π‘Žπ‘₯+𝑏) (π‘Ž . 1+0) γ€–=a cos 〗⁑(π‘Žπ‘₯+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) . (𝑑(𝑐π‘₯)/𝑑π‘₯ + 𝑑(𝑑)/𝑑π‘₯) γ€–=βˆ’sin 〗⁑(𝑐π‘₯+𝑏) . (𝑐 . 𝑑π‘₯/𝑑π‘₯ +0) γ€–=βˆ’sin 〗⁑(𝑐π‘₯+𝑏) (𝑐+0) γ€–=βˆ’c sin 〗⁑(𝑐π‘₯+𝑏) Now, 𝑑𝑦/𝑑π‘₯ = (𝑒^β€² 𝑣 βˆ’ γ€– 𝑣〗^β€² 𝑒)/𝑣^2 = (γ€–a cos 〗⁑〖(π‘Žπ‘₯+𝑏) .γ€– cos〗⁑〖 (𝑐π‘₯ + 𝑑)γ€— βˆ’(βˆ’γ€–sin 〗⁑〖(𝑐π‘₯ + 𝑑) . 𝑐〗 ) γ€— (sin⁑〖 (π‘Žπ‘₯+𝑏)γ€— ) )/(cos⁑〖 (𝑐π‘₯+𝑑)γ€— )^2 = (γ€–a cos 〗⁑〖(π‘Žπ‘₯+ 𝑏) .γ€– cos〗⁑〖 (𝑐π‘₯ + 𝑑)γ€— + 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) γ€– cos〗⁑〖 (𝑐π‘₯ + 𝑑)γ€—/cos^2⁑〖 (𝑐π‘₯+𝑑)γ€— + 𝑐 . γ€–sin 〗⁑(π‘Žπ‘₯+𝑏) . (γ€–sin 〗⁑〖(𝑐π‘₯ +𝑑)γ€— )/cos⁑〖 (𝑐π‘₯ + 𝑑)γ€— 1/cos⁑〖 (𝑐π‘₯ + 𝑑)γ€— = a cos (π‘Žπ‘₯+ 𝑏) 1/cos⁑〖 (𝑐π‘₯+𝑑)γ€— + 𝑐 . γ€–sin 〗⁑(π‘Žπ‘₯+𝑏) . (γ€–sin 〗⁑〖(𝑐π‘₯ +𝑑)γ€— )/cos⁑〖 (𝑐π‘₯ + 𝑑)γ€— 1/cos⁑〖 (𝑐π‘₯ + 𝑑)γ€— = 𝐚 𝒄𝒐𝒔 (𝒂𝒙+ 𝒃) .𝒔𝒆𝒄⁑〖(𝒄𝒙+𝒅)γ€— + 𝒄 . γ€–π’”π’Šπ’ 〗⁑(𝒂𝒙+𝒃).γ€–π­πšπ§ 〗⁑(𝒄𝒙+𝒅).𝒔𝒆𝒄⁑(𝒄𝒙+𝒅)

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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.