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Misc 10 - Differentiate xx + xa + ax + aa - Class 12 NCERT

Misc  10 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc  10 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Misc  10 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Misc  10 - Chapter 5 Class 12 Continuity and Differentiability - Part 5

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Misc 10 Differentiate w.r.t. x the function, π‘₯π‘₯ + π‘₯π‘Ž + π‘Ž^π‘₯+ π‘Žπ‘Ž, for some fixed π‘Ž >0 and π‘₯> 0Let 𝑦= π‘₯π‘₯ + π‘₯π‘Ž + π‘Ž^π‘₯+ π‘Žπ‘Ž And let u=π‘₯π‘₯ , 𝑣=π‘₯π‘Ž , 𝑀=π‘Ž^π‘₯ Now, π’š=𝒖+𝒗+π’˜+𝒂𝒂 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = 𝑑(𝑒 + 𝑣 + 𝑀 + π‘Žπ‘Ž)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(𝑒)/𝑑π‘₯ +𝑑(𝑣)/𝑑π‘₯+𝑑(𝑀)/𝑑π‘₯ + 𝑑(π‘Žπ‘Ž)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ +𝑑𝑣/𝑑π‘₯+𝑑𝑀/𝑑π‘₯ + 0 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ + 𝑑𝑀/𝑑π‘₯ Calculating 𝒅𝒖/𝒅𝒙 𝑒 =π‘₯^π‘₯ Taking log on both sides log⁑𝑒=log⁑〖π‘₯^π‘₯ γ€— log⁑𝑒=π‘₯ .log⁑π‘₯ (π‘Ž^π‘Ž 𝑖𝑠 π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑(log⁑𝑒 )/𝑑π‘₯ = 𝑑(π‘₯ .γ€– log〗⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑π‘₯ . 𝑑𝑒/𝑑𝑒 = 𝑑(π‘₯ .γ€– log〗⁑π‘₯ )/𝑑π‘₯ 𝑑(log⁑𝑒 )/𝑑𝑒 . 𝑑𝑒/𝑑π‘₯ = 𝑑π‘₯/𝑑π‘₯ . log⁑π‘₯ + (𝑑(γ€– log〗⁑π‘₯))/𝑑π‘₯. π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = log⁑π‘₯ + 1/π‘₯ . π‘₯ 1/𝑒 . 𝑑𝑒/𝑑π‘₯ = log⁑π‘₯ + 1 𝑑𝑒/𝑑π‘₯ = u (1+ log π‘₯)⁑ 𝒅𝒖/𝒅𝒙 = 𝒙^𝒙 (𝟏+ π’π’π’ˆ 𝒙)⁑ Calculating 𝒅𝒗/𝒅𝒙 𝑣=π‘₯^π‘Ž Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑣/𝑑π‘₯= 𝑑(π‘₯^π‘Ž )/𝑑π‘₯ 𝒅𝒗/𝒅𝒙= 𝒂𝒙^(𝒂 βˆ’πŸ) Calculating π’…π’˜/𝒅𝒙 𝑀=π‘Ž^π‘₯ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑀/𝑑π‘₯ = 𝑑(π‘Ž^π‘₯ )/𝑑π‘₯ 𝑑𝑀/𝑑π‘₯ = π‘Ž^π‘₯ .logβ‘π‘Ž Therefore, 𝑑𝑦/𝑑π‘₯ = 𝑑𝑒/𝑑π‘₯ + 𝑑𝑣/𝑑π‘₯ + 𝑑𝑀/𝑑π‘₯ = 𝒙^𝒙 (𝟏+ π’π’π’ˆ 𝒙) + 𝒂𝒙^(𝒂 βˆ’πŸ) + 𝒂^𝒙 .π’π’π’ˆβ‘π’‚

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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.