Ex 5.5, 9 - Differentiate x^sin x + (sin x)^cos x - Chapter 5 Class 12

Ex 5.5, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.5, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.5, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 5 Ex 5.5, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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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 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