Check sibling questions

Example 48 - Examples

Example 48 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


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 )

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.