Ex 5.3, 7 - Find dy/dx in sin2 y + cos xy = pi - Class 12

Ex 5.3, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.3, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.3, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Transcript

Ex 5.3, 7 Find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ in, sin2 ๐‘ฆ +cosโก ๐‘ฅ๐‘ฆ =๐œ‹ sin2 ๐‘ฆ +cosโก ๐‘ฅ๐‘ฆ =๐œ‹ Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ . (๐‘‘ (sin2 ๐‘ฆ + cosโก ๐‘ฅ๐‘ฆ))/๐‘‘๐‘ฅ = (๐‘‘ (๐œ‹))/๐‘‘๐‘ฅ (๐‘‘ (sin2 ๐‘ฆ))/๐‘‘๐‘ฅ + (๐‘‘ (cosโกใ€– ๐‘ฅใ€— ๐‘ฆ))/๐‘‘๐‘ฅ= 0 Calculating Derivative of sin2 ๐‘ฆ & cos (x๐‘ฆ) separately Calculating Derivative of ๐’”๐’Š๐’๐Ÿ ๐’š (๐‘‘ (sin2 ๐‘ฆ))/๐‘‘๐‘ฅ= (๐‘‘ (sin2 ๐‘ฆ))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ =(๐‘‘ (sin2(๐‘ฆ)))/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ =2 ๐‘ ๐‘–๐‘› ๐‘ฆ ร— (๐‘‘(sinโกใ€–๐‘ฆ)ใ€—)/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2 sinโกใ€–๐‘ฆ cosโกใ€–๐‘ฆ ร— ใ€— ใ€— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Calculating Derivative of ๐’„๐’๐’” (๐’™๐’š) (๐‘‘(cosโกใ€–(๐‘ฅ๐‘ฆ))ใ€—)/๐‘‘๐‘ฅ= โˆ’ sinโก(๐‘ฅ๐‘ฆ) ร— ๐‘‘/๐‘‘๐‘ฅ(๐‘ฅ๐‘ฆ) = โˆ’sinโก(๐‘ฅ๐‘ฆ)ร—(๐‘‘(๐‘ฅ)/๐‘‘๐‘ฅ.๐‘ฆ+๐‘‘(๐‘ฆ)/๐‘‘๐‘ฅ.๐‘ฅ) = โˆ’sinโก(๐‘ฅ๐‘ฆ) . (1.๐‘ฆ+๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = โˆ’sinโก(๐‘ฅ๐‘ฆ) (๐‘ฆ+๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = โˆ’sinโก(๐‘ฅ๐‘ฆ) . ๐‘ฆโˆ’sinโกใ€–(๐‘ฅ๐‘ฆ) . ๐‘ฅใ€— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’๐‘ฆ sinโกใ€– (๐‘ฅ๐‘ฆ)ใ€— โˆ’ sinโกใ€–(๐‘ฅ๐‘ฆ) . ๐‘ฅใ€— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Now, (๐‘‘ (๐‘ ๐‘–๐‘›2๐‘ฆ))/๐‘‘๐‘ฅ+(๐‘‘ (cosโกใ€–(๐‘ฅ๐‘ฆ)) ใ€—)/๐‘‘๐‘ฅ = 0 Putting values 2 sinโก๐‘ฆ.cosโกใ€–๐‘ฆ .ใ€— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ + (โˆ’ ๐‘ฆ sinโก(๐‘ฅ๐‘ฆ)โˆ’๐‘ฅ sinโก(๐‘ฅ๐‘ฆ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = 0 2 sinโก๐‘ฆ.cosโกใ€–๐‘ฆ .ใ€— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ’ ๐‘ฆ sinโก(๐‘ฅ๐‘ฆ) โˆ’ ๐‘ฅ sinโก(๐‘ฅ๐‘ฆ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 2 sinโก๐‘ฆ cosโกใ€–๐‘ฆ ใ€— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ’๐‘ฅ sinโก(๐‘ฅ๐‘ฆ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฆ sinโก(๐‘ฅ๐‘ฆ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (2 sinโก๐‘ฆ cosโก๐‘ฆ โˆ’ ๐‘ฅ sinโก(๐‘ฅ๐‘ฆ) = ๐‘ฆ sinโก(๐‘ฅ๐‘ฆ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘ฆ sinโกใ€–(๐‘ฅ๐‘ฆ)ใ€—)/(2 ใ€–sin ๐‘ฆใ€—โกใ€–cosโก๐‘ฆ โˆ’ ๐‘ฅ sinโก๐‘ฅ๐‘ฆ ใ€— ) ๐’…๐’š/๐’…๐’™ = (๐’š ๐’”๐’Š๐’โกใ€–(๐’™๐’š)ใ€—)/ใ€–๐’”๐’Š๐’ ๐Ÿ๐’šใ€—โกใ€–โˆ’ ๐’™ ๐’”๐’Š๐’โก๐’™๐’š ใ€—

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.