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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 43 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 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.