If y = 𝑒 π‘₯ 𝑠𝑖𝑛 2 π‘₯ + (sin π‘₯) π‘₯ , find dy/dx .

 

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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Question 30 If 𝑦=𝑒^(π‘₯ 〖𝑠𝑖𝑛〗^2⁑π‘₯ )+(𝑠𝑖𝑛⁑π‘₯ )^π‘₯, find 𝑑𝑦/𝑑π‘₯ . Let 𝑦=𝑒^(π‘₯ 〖𝑠𝑖𝑛〗^2⁑π‘₯ )+(sin⁑π‘₯ )^π‘₯ Let 𝑒 = 𝑒^(π‘₯ 〖𝑠𝑖𝑛〗^2⁑π‘₯ ) & 𝑣 =𝑒^(π‘₯ 〖𝑠𝑖𝑛〗^2⁑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 =𝑒^(π‘₯ 〖𝑠𝑖𝑛〗^2⁑π‘₯ ) Taking log both sides log⁑𝑒=log⁑〖𝑒^(π‘₯ 〖𝑠𝑖𝑛〗^2⁑π‘₯ ) γ€— log⁑𝑒=π‘₯ sin^2⁑π‘₯.log⁑𝑒 π’π’π’ˆβ‘π’–=𝒙 γ€–π’”π’Šπ’γ€—^πŸβ‘π’™ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. (𝑑(log⁑𝑒))/𝑑π‘₯ = 𝑑(π‘₯ . sin^2⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑π‘₯ . 𝑑𝑒/𝑑𝑒 = 𝑑(π‘₯ . sin^2⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(π‘₯ . sin^2⁑π‘₯ )/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑(π‘₯ . sin^2⁑π‘₯ )/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑π‘₯/𝑑π‘₯ . sin^2⁑π‘₯+ 𝑑(sin^2⁑π‘₯ )/𝑑π‘₯ Γ— π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 1 . sin^2⁑π‘₯ + (2 sin⁑π‘₯ cos⁑π‘₯ ) π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = sin^2⁑π‘₯+sin⁑2π‘₯ 𝑑𝑒/𝑑π‘₯ = u(sin^2⁑π‘₯+sin⁑2π‘₯ ) 𝒅𝒖/𝒅𝒙 = 𝒆^(𝒙 γ€–π’”π’Šπ’γ€—^πŸβ‘π’™ ) (γ€–π’”π’Šπ’γ€—^πŸβ‘π’™+π’”π’Šπ’β‘πŸπ’™ ) Calculating 𝒅𝒗/𝒅𝒙 𝑣 =(sin⁑π‘₯ )^π‘₯ Taking log both sides log⁑𝑣=log⁑〖(sin⁑π‘₯ )^π‘₯ γ€— π’π’π’ˆβ‘π’—=𝒙 .π₯𝐨𝐠⁑𝐬𝐒𝐧⁑〖𝒙 γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. (𝑑(log⁑𝑣))/𝑑π‘₯ = 𝑑(π‘₯.γ€– log〗⁑sin⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑣 )/𝑑π‘₯ . 𝑑𝑣/𝑑𝑣 = 𝑑(π‘₯ .γ€– log〗⁑sin⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑣 )/𝑑𝑣 . 𝑑𝑒/𝑑π‘₯ = 𝑑(π‘₯ .γ€– log〗⁑sin⁑π‘₯ )/𝑑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 𝑑(π‘₯ .γ€– log〗⁑sin⁑π‘₯ )/𝑑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 𝑑π‘₯/𝑑π‘₯ . log⁑sin⁑π‘₯ + 𝑑(log⁑sin⁑π‘₯ )/𝑑π‘₯ Γ— π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 1 . log⁑sin⁑π‘₯ + (1/sin⁑π‘₯ . 𝑑(sin⁑π‘₯ )/𝑑π‘₯) π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = log⁑sin⁑π‘₯ + (1/sin⁑π‘₯ . cos⁑π‘₯ )π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = log⁑sin⁑π‘₯ + π‘₯ (cos⁑π‘₯/sin⁑π‘₯ ) 𝑑𝑣/𝑑π‘₯ = v(log⁑sin⁑〖π‘₯+π‘₯ cot⁑π‘₯ γ€— ) 𝒅𝒗/𝒅𝒙 = (𝐬𝐒𝐧 𝒙)^𝒙 (π’π’π’ˆβ‘π’”π’Šπ’β‘γ€–π’™+𝒙 𝒄𝒐𝒕⁑𝒙 γ€— ) Now , 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ Putting value of 𝑑𝑒/𝑑π‘₯ & 𝑑𝑣/𝑑π‘₯ π’…π’š/𝒅𝒙 = 𝑒^(π‘₯ 〖𝑠𝑖𝑛〗^2⁑π‘₯ ) (sin^2⁑π‘₯+sin⁑2π‘₯ )+(𝐬𝐒𝐧⁑𝒙 )^𝒙 (𝒙 πœπ¨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.