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

Ex 5.5, 9 Differentiate the functions in, π‘₯^sin⁑π‘₯ + γ€–(sin⁑π‘₯)γ€—^cos⁑π‘₯ Let y = π‘₯^sin⁑π‘₯ + γ€–(sin⁑π‘₯)γ€—^cos⁑〖π‘₯ γ€— Let 𝑒 =π‘₯^sin⁑π‘₯ & 𝑣 =γ€–(sin⁑π‘₯)γ€—^cos⁑π‘₯ ∴ 𝑦 = 𝑒 + 𝑣 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = (𝑑 (𝑒 + 𝑣))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Calculating 𝒅𝒖/𝒅𝒙 𝑒 =π‘₯^sin⁑π‘₯ Taking log both sides log⁑𝑒=log⁑〖π‘₯^sin⁑π‘₯ γ€— log⁑𝑒= sin⁑π‘₯. log⁑〖 π‘₯γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. (𝑑(log⁑𝑒))/𝑑π‘₯ = 𝑑(sin⁑π‘₯. log⁑〖 π‘₯γ€— )/𝑑π‘₯ (𝑑(log⁑𝑒))/𝑑π‘₯ . 𝑑𝑒/𝑑𝑒 = 𝑑(sin⁑π‘₯. log⁑〖 π‘₯γ€— )/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(sin⁑π‘₯ . log⁑〖 π‘₯γ€— )/𝑑π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 . π‘™π‘œπ‘”β‘π‘Ž) Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(sin⁑π‘₯ )/𝑑π‘₯ . log⁑π‘₯ + 𝑑(log⁑π‘₯ )/𝑑π‘₯ . sin π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = cos⁑π‘₯ . log⁑π‘₯ + 1/π‘₯ . sin π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = cos⁑π‘₯ . log⁑π‘₯ + (sin π‘₯" " )/π‘₯ 𝑑𝑒/𝑑π‘₯ = u(cos⁑〖π‘₯ .log⁑〖π‘₯+ sin⁑π‘₯/π‘₯γ€— γ€— ) 𝑑𝑒/𝑑π‘₯ = π‘₯^sin⁑π‘₯ (cos⁑〖π‘₯ .log⁑〖π‘₯+ sin⁑π‘₯/π‘₯γ€— γ€— ). 𝑑𝑒/𝑑π‘₯ = π‘₯^sin⁑π‘₯ (γ€–sin⁑π‘₯/π‘₯ +cos〗⁑〖π‘₯ .log⁑π‘₯ γ€— ) Calculating 𝒅𝒗/𝒅𝒙 𝑣= (sin⁑π‘₯ )^cos⁑π‘₯ Taking log both sides log⁑𝑣 = log⁑〖 γ€–. sin〗⁑π‘₯γ€—^(cos π‘₯) log⁑𝑣=cos⁑π‘₯. log sin⁑π‘₯ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. (𝑑(log⁑𝑣))/𝑑π‘₯ = 𝑑(cos⁑π‘₯. log sin⁑π‘₯ )/𝑑π‘₯ (𝑑(log⁑𝑣))/𝑑π‘₯ . 𝑑𝑣/𝑑𝑣 = 𝑑(cos⁑π‘₯. log sin⁑π‘₯ )/𝑑π‘₯ (𝑑(log⁑𝑣))/𝑑𝑣 . 𝑑𝑣/𝑑π‘₯ = 𝑑(cos⁑π‘₯. log sin⁑π‘₯ )/𝑑π‘₯ (As π‘™π‘œπ‘”β‘(π‘Ž^𝑏) = 𝑏 . π‘™π‘œπ‘”β‘π‘Ž) 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 𝑑(cos⁑π‘₯. log sin⁑π‘₯ )/𝑑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 𝑑(cos⁑π‘₯ )/𝑑π‘₯ . log⁑sin⁑π‘₯ + 𝑑(log⁑sin⁑π‘₯ )/𝑑π‘₯ . cos⁑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = βˆ’sin⁑π‘₯ . log⁑sin⁑π‘₯ + (1/sin⁑π‘₯ . 𝑑(sin⁑π‘₯ )/𝑑π‘₯) . cos⁑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = βˆ’sin⁑π‘₯ . log⁑sin⁑π‘₯ + (1/sin⁑π‘₯ .cos⁑π‘₯ ) . cos⁑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = βˆ’sin⁑π‘₯ . log⁑sin⁑π‘₯ + (cot⁑π‘₯ ) . cos⁑π‘₯ 𝑑𝑣/𝑑π‘₯ = 𝑣(βˆ’sin⁑π‘₯ " " . log⁑sin⁑π‘₯" +" cot⁑π‘₯ ". " cos⁑π‘₯) 𝑑𝑣/𝑑π‘₯ = (sin⁑π‘₯ )^cos⁑π‘₯ (co𝑠⁑π‘₯ ". " co𝑑⁑π‘₯βˆ’sin⁑π‘₯ " " . log⁑sin⁑π‘₯) using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 Now 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Putting value of 𝑑𝑒/𝑑π‘₯ & 𝑑𝑣/𝑑π‘₯ π’…π’š/𝒅𝒙 = 𝒙^𝐬𝐒𝐧⁑𝒙 (𝐬𝐒𝐧⁑𝒙/𝒙+πœπ¨π¬β‘γ€–π’™ .π₯𝐨𝐠⁑𝒙 γ€— ) + (𝐬𝐒𝐧⁑𝒙 )^πœπ¨π¬β‘π’™ (πœπ¨π¬β‘π’™ .πœπ¨π­β‘π’™βˆ’π¬π’π§β‘π’™ π₯𝐨𝐠⁑𝐬𝐒𝐧⁑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.