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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 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.