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

 

If y = e^(x sin^2 x) + (sin x)^x, find dy/dx [with Video] - Teachoo

Question 30 - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 30 - CBSE Class 12 Sample Paper for 2021 Boards - Part 3
Question 30 - CBSE Class 12 Sample Paper for 2021 Boards - Part 4
Question 30 - CBSE Class 12 Sample Paper for 2021 Boards - Part 5 Question 30 - CBSE Class 12 Sample Paper for 2021 Boards - Part 6

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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‘₯ Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 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‘₯+π‘₯ sin⁑2π‘₯ 𝑑𝑒/𝑑π‘₯ = u(sin^2⁑π‘₯+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’™+𝒙 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 . 𝑑𝑣/𝑑π‘₯ = 𝑑(π‘₯ .γ€– log〗⁑sin⁑π‘₯ )/𝑑π‘₯ 1/𝑣 . 𝑑𝑣/𝑑π‘₯ = 𝑑(π‘₯ .γ€– log〗⁑sin⁑π‘₯ )/𝑑π‘₯ Using product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 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‘₯+π‘₯ sin⁑2π‘₯ )+(𝐬𝐒𝐧⁑𝒙 )^𝒙 (𝒙 πœπ¨π­β‘π’™+π₯𝐨𝐠⁑𝐬𝐒𝐧⁑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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.