Example 45 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Example 45 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Example 45 - Chapter 5 Class 12 Continuity and Differentiability - Part 5 Example 45 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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

Transcript

Example 45 Differentiate the following w.r.t. x. (ii) tan βˆ’1 (sin⁑π‘₯/( 1 +γ€– cos〗⁑〖π‘₯ γ€— )) Let 𝑓(π‘₯) = tan βˆ’1 (π’”π’Šπ’β‘π’™/( 1 +γ€– 𝒄𝒐𝒔〗⁑〖𝒙 γ€— )) 𝑓(π‘₯) = tan βˆ’1 ((𝟐 〖𝐬𝐒𝐧 〗⁑〖𝒙/πŸγ€— γ€– 𝐜𝐨𝐬 〗⁑〖𝒙/πŸγ€—)/( 1+ (𝟐 γ€–πœπ¨π¬γ€—^πŸβ‘γ€– 𝒙/πŸγ€— βˆ’ 𝟏))) = tan βˆ’1 ((2 γ€–sin 〗⁑〖π‘₯/2γ€— γ€– cos 〗⁑〖π‘₯/2 γ€—)/( 2 cos^2⁑〖 π‘₯/2γ€— )) = tan βˆ’1 ((2 γ€–sin 〗⁑〖π‘₯/2γ€—)/( 2 cos⁑〖π‘₯/2γ€— )) We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by πœƒ/2 sin ΞΈ = 2 π’”π’Šπ’β‘γ€–πœ½/πŸγ€— π’„π’π’”β‘γ€–πœ½/πŸγ€— and cos 2ΞΈ = 2cos2 ΞΈ – 1 Replacing ΞΈ by πœƒ/2 cos ΞΈ = 2cos2 𝜽/𝟐 – 1 = tan βˆ’1 (γ€–sin 〗⁑〖π‘₯/2γ€—/( cos⁑〖 π‘₯/2γ€— )) = tan βˆ’1 (γ€–tan 〗⁑〖π‘₯/2γ€— ) = 𝒙/𝟐 𝒇(𝒙) = 𝒙/𝟐 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑓’(π‘₯) = 1/2 (𝑑(π‘₯))/𝑑π‘₯ 𝒇’(𝒙) = 𝟏/𝟐 (As tan^(βˆ’1)⁑〖(tanβ‘πœƒ)γ€— =πœƒ) Example 45 (Method 2) Differentiate the following w.r.t. x. (ii) tan βˆ’1 (sin⁑π‘₯/( 1 +γ€– cos〗⁑〖π‘₯ γ€— )) Let 𝑓(π‘₯) = tan βˆ’1 (π’”π’Šπ’β‘π’™/( 1 +γ€– 𝒄𝒐𝒔〗⁑〖𝒙 γ€— )) Differentiating w.r.t x 𝑓^β€² (π‘₯) = 1/(1 + (π’”π’Šπ’β‘π’™/( 1 +γ€– 𝒄𝒐𝒔〗⁑〖𝒙 γ€— ))^2 ) (π’”π’Šπ’β‘π’™/( 1 +γ€– 𝒄𝒐𝒔〗⁑〖𝒙 γ€— ))^β€² = 1/(((1 + cos⁑π‘₯ )^2 + sin^2⁑π‘₯)/(1 + cos⁑π‘₯ )^2 ) (((π’”π’Šπ’β‘π’™ )^β€² (𝟏 + 𝒄𝒐𝒔 𝒙)βˆ’(𝟏 + 𝒄𝒐𝒔 𝒙)^β€² π’”π’Šπ’ 𝒙)/( (1 +γ€– 𝒄𝒐𝒔〗⁑𝒙 )^𝟐 )) = (1 + cos⁑π‘₯ )^2/((1 + cos⁑π‘₯ )^2 + sin^2⁑π‘₯ ) ((𝒄𝒐𝒔 𝒙(𝟏 + 𝒄𝒐𝒔 𝒙). βˆ’ (βˆ’ π’”π’Šπ’ 𝒙)π’”π’Šπ’ 𝒙)/( (1 +γ€– 𝒄𝒐𝒔〗⁑𝒙 )^𝟐 )) = (π‘π‘œπ‘  π‘₯ + 𝒄𝒐𝒔^𝟐 𝒙 + π’”π’Šπ’^𝟐 𝒙)/(1 + γ€–πœπ¨π¬γ€—^πŸβ‘π’™ + 2 cos⁑π‘₯ + γ€–π’”π’Šπ’γ€—^πŸβ‘π’™ ) = (𝒄𝒐𝒔 𝒙 +𝟏)/(1 + 1 + 2 cos⁑π‘₯ ) = (𝒄𝒐𝒔 𝒙 + 𝟏)/(2 + 2 cos⁑π‘₯ ) = (1 + 𝒄𝒐𝒔 𝒙)/(2(1 + cos⁑π‘₯) ) = 1/2

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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.