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

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  1. Class 12
  2. Important Question for exams Class 12
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Ex 5.2, 5 Differentiate the functions with respect to ๐‘ฅ sinโกใ€–(๐‘Ž๐‘ฅ+๐‘)ใ€—/cosโกใ€–(๐‘๐‘ฅ+๐‘‘)ใ€— Let ๐‘ฆ = sinโกใ€– (๐‘Ž๐‘ฅ+๐‘)ใ€—/cosโกใ€– (๐‘๐‘ฅ+๐‘‘)ใ€— Let ๐‘ข = sinโกใ€– (๐‘Ž๐‘ฅ+๐‘)ใ€— & ๐‘ฃ=cosโกใ€– (๐‘๐‘ฅ+๐‘‘)ใ€— ๐‘ฆ = ๐‘ข/๐‘ฃ We need to find derivative of ๐‘ฆ ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘ข/๐‘ฃ)^โ€ฒ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘ข^โ€ฒ ๐‘ฃ โˆ’ ใ€– ๐‘ฃใ€—^โ€ฒ ๐‘ข)/๐‘ฃ^2 Finding ๐’–โ€™ ๐‘ข=sinโกใ€– (๐‘Ž๐‘ฅ+๐‘)ใ€— Derivative of ๐‘ข ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ข/๐‘‘๐‘ฅ = ๐‘‘(sinโกใ€– (๐‘Ž๐‘ฅ+๐‘)ใ€— )/๐‘‘๐‘ฅ ใ€–=cos ใ€—โก(๐‘Ž๐‘ฅ+๐‘) . ๐‘‘(๐‘Ž๐‘ฅ + ๐‘)/๐‘‘๐‘ฅ ใ€–=cos ใ€—โก(๐‘Ž๐‘ฅ+๐‘) . (๐‘‘(๐‘Ž๐‘ฅ)/๐‘‘๐‘ฅ + ๐‘‘(๐‘)/๐‘‘๐‘ฅ) ใ€–=cos ใ€—โก(๐‘Ž๐‘ฅ+๐‘) . (๐‘Ž . ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ + ๐‘‘(๐‘)/๐‘‘๐‘ฅ) ใ€–=cos ใ€—โก(๐‘Ž๐‘ฅ+๐‘) (๐‘Ž . 1+0) ใ€–=a cos ใ€—โก(๐‘Ž๐‘ฅ+๐‘) Finding ๐’—โ€™ ๐‘ฃ=cosโกใ€– (๐‘๐‘ฅ+๐‘‘)ใ€— Derivative of ๐‘ฃ ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ = (๐‘‘(cosโกใ€– (๐‘๐‘ฅ + ๐‘‘)ใ€— )^โ€ฒ )/๐‘‘๐‘ฅ ใ€–=โˆ’si๐‘› ใ€—โก(๐‘๐‘ฅ+๐‘‘) . ๐‘‘(๐‘๐‘ฅ + ๐‘‘)/๐‘‘๐‘ฅ ใ€–=โˆ’sin ใ€—โก(๐‘๐‘ฅ+๐‘) . (๐‘‘(๐‘๐‘ฅ)/๐‘‘๐‘ฅ + ๐‘‘(๐‘‘)/๐‘‘๐‘ฅ) ใ€–=โˆ’sin ใ€—โก(๐‘๐‘ฅ+๐‘) . (๐‘ . ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ +0) ใ€–=โˆ’sin ใ€—โก(๐‘๐‘ฅ+๐‘) (๐‘+0) ใ€–=โˆ’c sin ใ€—โก(๐‘๐‘ฅ+๐‘) Now, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘ข^โ€ฒ ๐‘ฃ โˆ’ ใ€– ๐‘ฃใ€—^โ€ฒ ๐‘ข)/๐‘ฃ^2 = (ใ€–a cos ใ€—โกใ€–(๐‘Ž๐‘ฅ+๐‘) .ใ€– cosใ€—โกใ€– (๐‘๐‘ฅ + ๐‘‘)ใ€— โˆ’(โˆ’ใ€–sin ใ€—โกใ€–(๐‘๐‘ฅ + ๐‘‘) . ๐‘ใ€— ) ใ€— (sinโกใ€– (๐‘Ž๐‘ฅ+๐‘)ใ€— ) )/(cosโกใ€– (๐‘๐‘ฅ+๐‘‘)ใ€— )^2 = (ใ€–a cos ใ€—โกใ€–(๐‘Ž๐‘ฅ+ ๐‘) .ใ€– cosใ€—โกใ€– (๐‘๐‘ฅ + ๐‘‘)ใ€— + ๐‘ . ใ€–sin ใ€—โก(๐‘๐‘ฅ + ๐‘‘) ใ€— sinโกใ€– (๐‘Ž๐‘ฅ+๐‘)ใ€— )/cos^2โก(๐‘๐‘ฅ+๐‘‘) = (ใ€–a cos ใ€—โกใ€–(๐‘Ž๐‘ฅ+ ๐‘) .ใ€– cosใ€—โกใ€– (๐‘๐‘ฅ + ๐‘‘)ใ€— ใ€— )/cos^2โก(๐‘๐‘ฅ+๐‘‘) + (๐‘ . ใ€–sin ใ€—โกใ€–(๐‘๐‘ฅ + ๐‘‘) ใ€—. sinโกใ€– (๐‘Ž๐‘ฅ+๐‘)ใ€— )/cos^2โก(๐‘๐‘ฅ+๐‘‘) = a cos (๐‘Ž๐‘ฅ+ ๐‘) ใ€– cosใ€—โกใ€– (๐‘๐‘ฅ + ๐‘‘)ใ€—/cos^2โกใ€– (๐‘๐‘ฅ+๐‘‘)ใ€— + ๐‘ . ใ€–sin ใ€—โก(๐‘Ž๐‘ฅ+๐‘) . (ใ€–sin ใ€—โกใ€–(๐‘๐‘ฅ +๐‘‘)ใ€— )/cosโกใ€– (๐‘๐‘ฅ + ๐‘‘)ใ€— 1/cosโกใ€– (๐‘๐‘ฅ + ๐‘‘)ใ€— = a cos (๐‘Ž๐‘ฅ+ ๐‘) 1/cosโกใ€– (๐‘๐‘ฅ+๐‘‘)ใ€— + ๐‘ . ใ€–sin ใ€—โก(๐‘Ž๐‘ฅ+๐‘) . (ใ€–sin ใ€—โกใ€–(๐‘๐‘ฅ +๐‘‘)ใ€— )/cosโกใ€– (๐‘๐‘ฅ + ๐‘‘)ใ€— 1/cosโกใ€– (๐‘๐‘ฅ + ๐‘‘)ใ€— = ๐š ๐’„๐’๐’” (๐’‚๐’™+ ๐’ƒ) .๐’”๐’†๐’„โกใ€–(๐’„๐’™+๐’…)ใ€— + ๐’„ . ใ€–๐’”๐’Š๐’ ใ€—โก(๐’‚๐’™+๐’ƒ).ใ€–๐ญ๐š๐ง ใ€—โก(๐’„๐’™+๐’…).๐’”๐’†๐’„โก(๐’„๐’™+๐’…)

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