Example 43 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 43 Differentiate 〖𝑠𝑖𝑛〗^2 𝑥 𝑤.𝑟.𝑡. 𝑒^(cos𝑥 ". " )Let 𝑢 = 〖𝑠𝑖𝑛〗^2 𝑥 & 𝑣 =𝑒^(cos𝑥 ) We need to differentiate 𝑢 𝑤.𝑟.𝑡. 𝑣 . i.e., 𝑑𝑢/𝑑𝑣 Here, 𝒅𝒖/𝒅𝒗 = (𝒅𝒖/𝒅𝒙)/(𝒅𝒗/𝒅𝒙) Calculating 𝒅𝒖/𝒅𝒙 𝑢 = 〖𝑠𝑖𝑛〗^2 𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑢/𝑑𝑥 = 𝑑(〖𝑠𝑖𝑛〗^2 𝑥)/𝑑𝑥 𝑑𝑢/𝑑𝑥 = 2 sin𝑥 . 𝑑(sin𝑥 )/𝑑𝑥 𝑑𝑢/𝑑𝑥 = 𝟐 𝒔𝒊𝒏𝒙 . 𝐜𝐨𝐬𝒙 Calculating 𝒅𝒗/𝒅𝒙 𝑣 =𝑒^(cos𝑥 ) Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑣/𝑑𝑥 = 𝑑(𝑒^(cos𝑥 ) )/𝑑𝑥 𝑑𝑣/𝑑𝑥 = 𝑒^(cos𝑥 ) . 𝑑(cos𝑥 )/𝑑𝑥 𝑑𝑣/𝑑𝑥 = 𝑒^(cos𝑥 ) . (−sin𝑥 ) 𝑑𝑣/𝑑𝑥 = −𝒔𝒊𝒏𝒙. 𝒆^(𝒄𝒐𝒔𝒙 )Therefore 𝑑𝑢/𝑑𝑣 = (𝑑𝑢/𝑑𝑥)/(𝑑𝑣/𝑑𝑥) = (2 sin𝑥" ." cos𝑥)/(−sin𝑥 . 𝑒^(cos𝑥 ) ) = (−𝟐"." 𝒄𝒐𝒔𝒙)/𝒆^(𝒄𝒐𝒔𝒙 )
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Example 43 You are here
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About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo