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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


Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.