Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

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

Transcript

Example 48 Differentiate 〖𝑠𝑖𝑛〗^2 π‘₯ 𝑀.π‘Ÿ.𝑑. 𝑒^(cos⁑π‘₯ ". " )Let 𝑒 = 〖𝑠𝑖𝑛〗^2 π‘₯ & 𝑣 =𝑒^(cos⁑π‘₯ ) We need to differentiate 𝑒 𝑀.π‘Ÿ.𝑑. 𝑣 . i.e., 𝑑𝑒/𝑑𝑣 Here, 𝒅𝒖/𝒅𝒗 = (𝒅𝒖/𝒅𝒙)/(𝒅𝒗/𝒅𝒙) Calculating 𝒅𝒖/𝒅𝒙 𝑒 = 〖𝑠𝑖𝑛〗^2 π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑒/𝑑π‘₯ = 𝑑(〖𝑠𝑖𝑛〗^2 π‘₯)/𝑑π‘₯ 𝑑𝑒/𝑑π‘₯ = 2 sin⁑π‘₯ . 𝑑(sin⁑π‘₯ )/𝑑π‘₯ 𝑑𝑒/𝑑π‘₯ = 𝟐 π’”π’Šπ’β‘π’™ . πœπ¨π¬β‘π’™ Calculating 𝒅𝒗/𝒅𝒙 𝑣 =𝑒^(cos⁑π‘₯ ) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑣/𝑑π‘₯ = 𝑑(𝑒^(cos⁑π‘₯ ) )/𝑑π‘₯ 𝑑𝑣/𝑑π‘₯ = 𝑒^(cos⁑π‘₯ ) . 𝑑(cos⁑π‘₯ )/𝑑π‘₯ 𝑑𝑣/𝑑π‘₯ = 𝑒^(cos⁑π‘₯ ) . (βˆ’sin⁑π‘₯ ) 𝑑𝑣/𝑑π‘₯ = βˆ’π’”π’Šπ’β‘π’™. 𝒆^(𝒄𝒐𝒔⁑𝒙 ) Therefore 𝑑𝑒/𝑑𝑣 = (𝑑𝑒/𝑑π‘₯)/(𝑑𝑣/𝑑π‘₯) = (2 sin⁑π‘₯" ." cos⁑π‘₯)/(βˆ’sin⁑π‘₯ . 𝑒^(cos⁑π‘₯ ) ) = (βˆ’πŸ"." 𝒄𝒐𝒔⁑𝒙)/𝒆^(𝒄𝒐𝒔⁑𝒙 )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.