# Ex 5.5, 9 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.5, 9 (Method 1) Differentiate the functions in, 𝑥 sin𝑥+ ( sin𝑥) cos𝑥 Let y = 𝑥 sin𝑥+ ( sin𝑥) cos𝑥 Let 𝑢 = 𝑥 sin𝑥 & 𝑣 = ( sin𝑥) cos𝑥 ∴ 𝑦 = 𝑢 + 𝑣 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦𝑑𝑥 = 𝑑 (𝑢 + 𝑣)𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑢𝑑𝑥 + 𝑑𝑣𝑑𝑥 Calculating 𝒅𝒖𝒅𝒙 𝑢 = 𝑥 sin𝑥 Taking log both sides log𝑢= log 𝑥 sin𝑥 log𝑢= sin𝑥. log 𝑥 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑( log𝑢)𝑑𝑥 = 𝑑 sin𝑥. log 𝑥𝑑𝑥 𝑑( log𝑢)𝑑𝑥 . 𝑑𝑢𝑑𝑢 = 𝑑 sin𝑥. log 𝑥𝑑𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = 𝑑 sin𝑥 . log 𝑥𝑑𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = 𝑑 sin𝑥𝑑𝑥 . log𝑥 + 𝑑 log𝑥𝑑𝑥 . sin 𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = cos𝑥 . log𝑥 + 1𝑥 . sin 𝑥 1𝑢 . 𝑑𝑢𝑑𝑥 = cos𝑥 . log𝑥 + sin 𝑥 𝑥 𝑑𝑢𝑑𝑥 = u cos𝑥 . log𝑥+ sin𝑥𝑥 𝑑𝑢𝑑𝑥 = 𝑥 sin𝑥 cos𝑥 . log𝑥+ sin𝑥𝑥. 𝑑𝑢𝑑𝑥 = 𝑥 sin𝑥 sin𝑥𝑥 +cos𝑥 . log𝑥 Calculating 𝒅𝒗𝒅𝒙 𝑣= sin𝑥 cos𝑥 Taking log both sides log𝑣 = log . sin𝑥cos 𝑥 log𝑣= cos𝑥. log sin𝑥 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑( log𝑣)𝑑𝑥 = 𝑑 cos𝑥. log sin𝑥𝑑𝑥 𝑑( log𝑣)𝑑𝑥 . 𝑑𝑣𝑑𝑣 = 𝑑 cos𝑥. log sin𝑥𝑑𝑥 𝑑( log𝑣)𝑑𝑣 . 𝑑𝑣𝑑𝑥 = 𝑑 cos𝑥. log sin𝑥𝑑𝑥 1𝑣 . 𝑑𝑣𝑑𝑥 = 𝑑 cos𝑥. log sin𝑥𝑑𝑥 1𝑣 . 𝑑𝑣𝑑𝑥 = 𝑑 cos𝑥𝑑𝑥 . logsin𝑥 + 𝑑 log sin𝑥𝑑𝑥 . cos𝑥 1𝑣 . 𝑑𝑣𝑑𝑥 = − sin𝑥 . logsin𝑥 + 1 sin𝑥 . 𝑑 sin𝑥𝑑𝑥 . cos𝑥 1𝑣 . 𝑑𝑣𝑑𝑥 = − sin𝑥 . logsin𝑥 + 1 sin𝑥 . cos𝑥 . cos𝑥 1𝑣 . 𝑑𝑣𝑑𝑥 = − sin𝑥 . logsin𝑥 + cot𝑥 . cos𝑥 𝑑𝑣𝑑𝑥 = 𝑣 − sin𝑥 . logsin𝑥 + cot𝑥. cos𝑥 𝑑𝑣𝑑𝑥 = sin𝑥 cos𝑥 co𝑠𝑥. co𝑡𝑥− sin𝑥 . logsin𝑥 Now 𝑑𝑦𝑑𝑥 = 𝑑𝑢𝑑𝑥 + 𝑑𝑣𝑑𝑥 Putting value of 𝑑𝑢𝑑𝑥 & 𝑑𝑣𝑑𝑥 𝒅𝒚𝒅𝒙 = 𝒙 𝐬𝐢𝐧𝒙 𝐬𝐢𝐧𝒙𝒙+ 𝐜𝐨𝐬𝒙 . 𝐥𝐨𝐠𝒙 + 𝐬𝐢𝐧𝒙 𝐜𝐨𝐬𝒙 𝒄𝒐𝒔𝒙. 𝒄𝒐𝒕𝒙− 𝒔𝒊𝒏𝒙 . 𝒍𝒐𝒈𝒔𝒊𝒏

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Davneet Singh

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.