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Derivative of cot-1 x (cot inverse x) - Teachoo  [with Video]

Derivative of cot-1 x (cot inverse x) - Part 2

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Derivative of 〖𝒄𝒐𝒕〗^(βˆ’πŸ) 𝒙 𝑓 (π‘₯)=γ€–π‘π‘œπ‘‘γ€—^(βˆ’1) π‘₯ Let π’š= 〖𝒄𝒐𝒕〗^(βˆ’πŸ) 𝒙 cot⁑〖𝑦=π‘₯γ€— 𝒙=πœπ¨π­β‘γ€–π’š γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑π‘₯/𝑑π‘₯ = (𝑑 (cot⁑𝑦 ))/𝑑π‘₯ 1 = (𝑑 (cot⁑𝑦 ))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 1 = (𝑑 (cot⁑𝑦 ))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ 1 = βˆ’πœπ¨γ€–π¬πžπœγ€—^𝟐 π’š . 𝑑𝑦/𝑑π‘₯ 1 = βˆ’(𝟏 +π’„π’π’•πŸπ’š) 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/(1 + γ€–πœπ¨π­γ€—^πŸβ‘π’š ) Putting π‘π‘œπ‘‘β‘π‘¦ = π‘₯ 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/(𝒙^𝟐 + 𝟏) Hence (𝒅(γ€–πœπ¨π­γ€—^(βˆ’πŸ)⁑〖𝒙)γ€—)/𝒅𝒙 = (βˆ’πŸ)/(𝒙^𝟐 + 𝟏) (𝐴𝑠 γ€– π‘π‘œπ‘ π‘’π‘γ€—^2⁑〖𝑦= γ€–1+γ€—β‘γ€–π‘π‘œπ‘‘γ€—^2⁑𝑦 γ€—) As 𝑦 = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1) π‘₯ So, π’„π’π’•β‘π’š = 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.