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Example 27 - Find derivative of f(x) = tan-1 x - Class 12

Example 27 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
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Example 27 Find the derivative of f given by f (x) = tan–1 π‘₯ assuming it exists. 𝑓 (π‘₯)=γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ Let π’š= 〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒙 tan⁑〖𝑦=π‘₯γ€— 𝒙=π­πšπ§β‘γ€–π’š γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑π‘₯/𝑑π‘₯ = (𝑑 (tan⁑𝑦 ))/𝑑π‘₯ 1 = (𝑑 (tan⁑𝑦 ))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 1 = (𝑑 (tan⁑𝑦 ))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ 1 = γ€–π¬πžπœγ€—^𝟐 π’š . 𝑑𝑦/𝑑π‘₯ 1 = (𝟏 + π’•π’‚π’πŸπ’š) 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/(1 + γ€–π­πšπ§γ€—^πŸβ‘π’š ) Putting π‘‘π‘Žπ‘›β‘π‘¦ = π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/(1 + 𝒙^𝟐 ) Hence (𝒅(γ€–π­πšπ§γ€—^(βˆ’πŸ)⁑〖𝒙)γ€—)/𝒅𝒙 = 𝟏/(𝟏 + 𝒙^𝟐 ) As 𝑦 = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ So, π’•π’‚π’β‘π’š = 𝒙 Derivative of 〖𝒄𝒐𝒔〗^(βˆ’πŸ) 𝒙 𝑓 (π‘₯)=γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ Let π’š= 〖𝒄𝒐𝒔〗^(βˆ’πŸ) 𝒙 cos⁑〖𝑦=π‘₯γ€— 𝒙=πœπ¨π¬β‘γ€–π’š γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑π‘₯/𝑑π‘₯ = (𝑑 (cos⁑𝑦 ))/𝑑π‘₯ 1 = (𝑑 (cos⁑𝑦 ))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 1 = (𝑑 (cos⁑𝑦 ))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ 1 = (βˆ’sin⁑𝑦) 𝑑𝑦/𝑑π‘₯ (βˆ’1)/sin⁑𝑦 =𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/π’”π’Šπ’β‘π’š 𝑑𝑦/𝑑π‘₯= (βˆ’1)/√(𝟏 βˆ’ 〖𝒄𝒐𝒔〗^𝟐 π’š) Putting π‘π‘œπ‘ β‘γ€–π‘¦=π‘₯γ€— 𝑑𝑦/𝑑π‘₯= (βˆ’1)/√(1 βˆ’ 𝒙^𝟐 ) Hence, (𝒅(〖𝒄𝒐𝒔〗^(βˆ’πŸ) 𝒙" " ))/𝒅𝒙 = (βˆ’πŸ)/√(𝟏 βˆ’ 𝒙^𝟐 ) "We know that" 〖𝑠𝑖𝑛〗^2 πœƒ+γ€–π‘π‘œπ‘ γ€—^2 πœƒ=1 〖𝑠𝑖𝑛〗^2 πœƒ=1βˆ’γ€–π‘π‘œπ‘ γ€—^2 πœƒ π’”π’Šπ’β‘πœ½=√(πŸβˆ’γ€–π’„π’π’”γ€—^𝟐 𝜽) " " As 𝑦 = γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ So, π’„π’π’”β‘π’š = 𝒙 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 Derivative of 〖𝒔𝒆𝒄〗^(βˆ’πŸ) 𝒙 𝑓 (π‘₯)=〖𝑠𝑒𝑐〗^(βˆ’1) π‘₯ Let π’š= 〖𝒔𝒆𝒄〗^(βˆ’πŸ) 𝒙 sec⁑〖𝑦=π‘₯γ€— 𝒙=π¬πžπœβ‘γ€–π’š γ€— Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑π‘₯/𝑑π‘₯ = (𝑑 (sec⁑𝑦 ))/𝑑π‘₯ 1 = (𝑑 (sec⁑𝑦 ))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 1 = (𝑑 (sec⁑𝑦 ))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ 1 = π’•π’‚π’β‘π’š .π’”π’†π’„β‘π’š. 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/(π’•π’‚π’β‘π’š .γ€– sec〗⁑𝑦 ) 𝑑𝑦/𝑑π‘₯ = 1/((√(γ€–π¬πžπœγ€—^πŸβ‘π’š βˆ’ 𝟏)) .γ€– sec〗⁑𝑦 ) Putting value of 𝑠𝑒𝑐⁑𝑦 = π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/((√(π‘₯^2 βˆ’ 1 ) ) . π‘₯) 𝑑𝑦/𝑑π‘₯ = 1/(π‘₯ √(π‘₯^2 βˆ’ 1 ) ) Hence 𝒅(〖𝒔𝒆𝒄〗^(β€“πŸ) 𝒙)/𝒅𝒙 = 𝟏/(𝒙 √(𝒙^𝟐 βˆ’ 𝟏 ) ) As tan2 ΞΈ = sec2 ΞΈ – 1, tan ΞΈ = √("sec2 ΞΈ – 1" ) As 𝑦 = 〖𝑠𝑒𝑐〗^(βˆ’1) π‘₯ So, π’”π’†π’„β‘π’š = 𝒙Derivative of 〖𝒄𝒐𝒔𝒆𝒄〗^(βˆ’πŸ) 𝒙 𝑓 (π‘₯)=γ€–π‘π‘œπ‘ π‘’π‘γ€—^(βˆ’1) π‘₯ Let π’š= 〖𝒄𝒐𝒔𝒆𝒄〗^(βˆ’πŸ) 𝒙 cosec⁑〖𝑦=π‘₯γ€— 𝒙=πœπ¨π¬πžπœβ‘γ€–π’š γ€— 1 = (𝑑 (cosec⁑𝑦 ))/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯ 1 = βˆ’cosec⁑𝑦 .cot⁑𝑦 . 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/(γ€–βˆ’cosec〗⁑𝑦 .π’„π’π’•β‘π’š ) 𝑑𝑦/𝑑π‘₯ = 1/(γ€–βˆ’cosec〗⁑𝑦 . √(γ€–πœπ¨π’”π’†π’„γ€—^πŸβ‘π’š βˆ’ 𝟏)) Putting value of π‘π‘œπ‘ π‘’π‘β‘π‘¦ = π‘₯ 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/(π‘₯ √(π‘₯^2 βˆ’ 1 ) ) Hence 𝒅(〖𝒄𝒐𝒔𝒆𝒄〗^(β€“πŸ) 𝒙)/𝒅𝒙 = (βˆ’πŸ)/(𝒙 √(𝒙^𝟐 βˆ’ 𝟏 ) ) As cot2 ΞΈ = cosec2 ΞΈ – 1, cot ΞΈ = √("cosec2 ΞΈ – 1" ) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑π‘₯/𝑑π‘₯ = (𝑑 (cosec⁑𝑦 ))/𝑑π‘₯ 1 = (𝑑 (cosec⁑𝑦 ))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 As cot2 ΞΈ = cosec2 ΞΈ – 1, cot ΞΈ = √("cosec2 ΞΈ – 1" ) As 𝑦 = co〖𝑠𝑒𝑐〗^(βˆ’1) π‘₯ So, coπ’”π’†π’„β‘π’š = 𝒙

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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.