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Example 33 - Find dy/dx if y^x + x^y + x^x = a^b - Teachoo

Example 33 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Example 33 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Example 33 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Example 33 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Example 33 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Example 33 - Chapter 5 Class 12 Continuity and Differentiability - Part 7
Example 33 - Chapter 5 Class 12 Continuity and Differentiability - Part 8
Example 33 - Chapter 5 Class 12 Continuity and Differentiability - Part 9

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Example 33 Find 𝑑𝑦/𝑑π‘₯ , if 𝑦^π‘₯+π‘₯^𝑦+π‘₯^π‘₯=π‘Ž^𝑏. Let u = 𝑦π‘₯, v = π‘₯𝑦 & w = π‘₯^π‘₯ Now, 𝒖 + 𝒗 + π’˜ = 𝒂^𝒃 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑 (𝑒 + 𝑣 + 𝑀))/𝑑π‘₯ = (𝑑(π‘Ž^𝑏))/𝑑π‘₯ (𝑑(𝑒))/𝑑π‘₯ + (𝑑(𝑣))/𝑑π‘₯ + (𝑑(𝑀))/𝑑π‘₯ = 0 We will calculate derivative of u, v & w separately . (As π‘Ž^𝑏 is constant) Finding Derivative of 𝒖 . 𝑒 = 𝑦^π‘₯ Taking log both sides log⁑𝑒=log⁑〖 (𝑦^π‘₯)" " γ€— log⁑𝑒=γ€–π‘₯ . log〗⁑𝑦" " Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑(log⁑𝑒))/𝑑π‘₯ = (𝑑(π‘₯ . log⁑𝑦))/𝑑π‘₯ (𝑑(log⁑𝑒))/𝑑π‘₯ (𝑑𝑒/𝑑𝑒) = 𝑑(π‘₯.log⁑𝑦 )/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = (𝑑 (π‘₯ . log⁑𝑦 ))/𝑑π‘₯ (𝐴𝑠 log⁑〖(π‘Ž^𝑏)γ€—=𝑏 logβ‘π‘Ž) By product Rule (uv)’ = u’v + v’u 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑π‘₯/𝑑π‘₯ . log⁑𝑦 + (𝑑(log⁑𝑦))/𝑑π‘₯ . π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = 1 . log⁑𝑦 + π‘₯. 𝑑(log⁑𝑦 )/𝑑π‘₯ . 𝑑𝑦/𝑑𝑦 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = log⁑𝑦 + π‘₯. 𝑑(log⁑𝑦 )/𝑑π‘₯ . 𝑑𝑦/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = log⁑𝑦 + π‘₯. 1/𝑦 . 𝑑𝑦/𝑑π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = log⁑𝑦 + π‘₯/𝑦 . 𝑑𝑦/𝑑π‘₯ 𝑑𝑒/𝑑π‘₯ = 𝑒 (log⁑𝑦 "+ " π‘₯/𝑦 " " 𝑑𝑦/𝑑π‘₯) 𝒅𝒖/𝒅𝒙 = π’š^𝒙 (π’π’π’ˆβ‘π’š "+ " 𝒙/π’š " " π’…π’š/𝒅𝒙) Finding derivative of v v = xy Taking log both sides log⁑𝑣=log⁑〖 (π‘₯^𝑦)" " γ€— log⁑𝑣=〖𝑦. log〗⁑π‘₯" " Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑(log⁑𝑣))/𝑑π‘₯ = (𝑑(𝑦 . log⁑π‘₯))/𝑑π‘₯ (𝑑(log⁑𝑣))/𝑑π‘₯ (𝑑𝑣/𝑑π‘₯) = 𝑑(〖𝑦 log〗⁑π‘₯ )/𝑑π‘₯ 1/𝑣 (𝑑𝑣/𝑑π‘₯) = ( 𝑑(〖𝑦 log〗⁑π‘₯ ))/𝑑π‘₯ By product Rule (uv)’ = u’v + v’u 1/𝑣 (𝑑𝑣/𝑑π‘₯) = ( 𝑑(𝑦))/𝑑π‘₯ . log⁑π‘₯ + (𝑑 (log⁑π‘₯))/𝑑π‘₯ . 𝑦 1/𝑣 (𝑑𝑣/𝑑π‘₯) = ( 𝑑(𝑦))/𝑑π‘₯ . log⁑π‘₯ + (𝑑 (log⁑π‘₯))/𝑑π‘₯ . 𝑦 1/𝑣 (𝑑𝑣/𝑑π‘₯) = ( 𝑑𝑦)/𝑑π‘₯ . log⁑π‘₯ + 1/π‘₯ . 𝑦 1/𝑣 (𝑑𝑣/𝑑π‘₯) = ( 𝑑𝑦)/𝑑π‘₯ log⁑π‘₯ + 𝑦/π‘₯ 𝑑𝑣/𝑑π‘₯ = v (log ( 𝑑𝑦)/𝑑π‘₯ π‘₯+𝑦/π‘₯) Putting values of 𝑣 = π‘₯^𝑦 𝒅𝒗/𝒅𝒙 = 𝒙^π’š (π’…π’š/𝒅𝒙 π’π’π’ˆβ‘γ€–π’™+ π’š/𝒙〗 ) Calculating derivative of π’˜ 𝑀 = π‘₯^π‘₯ Taking log both sides log⁑𝑀=log⁑〖 (π‘₯^π‘₯)" " γ€— log⁑𝑀=γ€–π‘₯. log〗⁑π‘₯" " Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑(log⁑𝑀))/𝑑π‘₯ = (𝑑(π‘₯ . log⁑π‘₯))/𝑑π‘₯ (𝑑(log⁑𝑀))/𝑑π‘₯ (𝑑𝑀/𝑑𝑀) = 𝑑(π‘₯ log⁑π‘₯ )/𝑑π‘₯ (𝑑(log⁑𝑀))/𝑑𝑀 . 𝑑𝑀/𝑑π‘₯ = 𝑑(π‘₯ log⁑π‘₯ )/𝑑π‘₯ (𝐴𝑠 log⁑〖(π‘Ž^𝑏)γ€—=𝑏 logβ‘π‘Ž) 1/𝑀 . 𝑑𝑀/𝑑π‘₯ = 𝑑(π‘₯ log⁑π‘₯ )/𝑑π‘₯ log⁑𝑀=γ€–π‘₯. log〗⁑π‘₯" " Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑(log⁑𝑀))/𝑑π‘₯ = (𝑑(π‘₯ . log⁑π‘₯))/𝑑π‘₯ (𝑑(log⁑𝑀))/𝑑π‘₯ (𝑑𝑀/𝑑𝑀) = 𝑑(π‘₯ log⁑π‘₯ )/𝑑π‘₯ (𝑑(log⁑𝑀))/𝑑𝑀 . 𝑑𝑀/𝑑π‘₯ = 𝑑(π‘₯ log⁑π‘₯ )/𝑑π‘₯ 1/𝑀 . 𝑑𝑀/𝑑π‘₯ = 𝑑(π‘₯ log⁑π‘₯ )/𝑑π‘₯ By product Rule (uv)’ = u’v + v’u 1/𝑀 (𝑑𝑀/𝑑π‘₯) = ( 𝑑(π‘₯))/𝑑π‘₯ . log⁑π‘₯ + (𝑑 (log⁑π‘₯))/𝑑π‘₯ . π‘₯ 1/𝑀 (𝑑𝑀/𝑑π‘₯) = 1 . log⁑π‘₯ + 1/π‘₯ . π‘₯ 1/𝑀 (𝑑𝑀/𝑑π‘₯) = (log⁑〖π‘₯+1γ€—) 𝑑𝑀/𝑑π‘₯ = 𝑀(log⁑〖π‘₯+1γ€—) π’…π’˜/𝒅𝒙 = 𝒙^𝒙 (π’π’π’ˆβ‘γ€–π’™+πŸγ€— ) From (1) 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ + 𝑑𝑀/𝑑π‘₯ = 0 Putting values from (2), (3) & (4) (𝑦^π‘₯ log⁑〖𝑦+𝑦^(π‘₯βˆ’1). π‘₯ 𝑑𝑦/𝑑π‘₯ γ€— ) + (π‘₯^𝑦 log⁑〖π‘₯.𝑑𝑦/𝑑π‘₯+π‘₯^𝑦.𝑦/π‘₯ γ€— ) + (π‘₯^π‘₯ (log⁑〖π‘₯+1γ€—))=0 (𝑦^π‘₯ log⁑〖𝑦+π‘₯^𝑦. 𝑦/π‘₯+π‘₯^π‘₯ (log⁑〖π‘₯+1γ€—)γ€— ) + (𝑦^(π‘₯βˆ’1) .⁑〖π‘₯ 𝑑𝑦/𝑑π‘₯+π‘₯^𝑦 log⁑〖π‘₯ 𝑑𝑦/𝑑π‘₯γ€— γ€— ) = 0 (𝑦^(π‘₯βˆ’1) .⁑〖π‘₯ 𝑑𝑦/𝑑π‘₯+π‘₯^𝑦 log⁑〖𝑦 𝑑𝑦/𝑑π‘₯γ€— γ€— ) = βˆ’ (𝑦^π‘₯ log⁑〖𝑦+π‘₯^𝑦. 𝑦/π‘₯+π‘₯^π‘₯ (log⁑〖π‘₯+1γ€—)γ€— ) (𝑦^(π‘₯βˆ’1) .⁑〖π‘₯ +π‘₯^𝑦 log⁑〖π‘₯ γ€— γ€— ) 𝑑𝑦/𝑑π‘₯ = βˆ’ (𝑦^π‘₯ log⁑〖𝑦+π‘₯^𝑦. 𝑦/π‘₯+π‘₯^π‘₯ (log⁑〖π‘₯+1γ€—)γ€— ) 𝑑𝑦/𝑑π‘₯ = "βˆ’" (𝑦^π‘₯ π‘™π‘œπ‘”β‘γ€–π‘¦ + π‘₯^𝑦. 𝑦/π‘₯ + π‘₯^π‘₯ (1 + π‘™π‘œπ‘”β‘π‘₯)γ€— )/((γ€–π‘₯𝑦〗^(π‘₯βˆ’1) +⁑〖π‘₯^𝑦 π‘™π‘œπ‘”β‘γ€–π‘₯ γ€— γ€—)) π’…π’š/𝒅𝒙 = "βˆ’" (π’š^𝒙 π’π’π’ˆβ‘γ€–π’š + 𝒙^(π’š βˆ’ 𝟏) π’š + 𝒙^𝒙 (𝟏 + π’π’π’ˆβ‘π’™)γ€— )/((γ€–π’™π’šγ€—^(π’™βˆ’πŸ) +⁑〖𝒙^π’š π’π’π’ˆβ‘γ€–π’™ γ€— γ€—))

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.