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

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Example 27 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Concept wise

Transcript

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.