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

Transcript

Example 45 (Method 1) Differentiate the following 𝑀.π‘Ÿ.𝑑. π‘₯. (i) cos^(βˆ’1) (sin⁑π‘₯) Let 𝑓(π‘₯) = cos^(βˆ’1) (sin⁑π‘₯) 𝑓(π‘₯) = cos^(βˆ’1) (γ€–cos 〗⁑(πœ‹/2 βˆ’π‘₯) ) 𝒇(𝒙) = 𝝅/𝟐 βˆ’π’™ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑓’(π‘₯) = (𝑑 (πœ‹/2))/𝑑π‘₯ βˆ’ (𝑑(π‘₯))/𝑑π‘₯ 𝑓’(π‘₯) = 0 βˆ’ 1 𝒇’(𝒙) = βˆ’ 1 (𝐴𝑠 γ€– 𝑠𝑖𝑛 πœƒ 〗⁑〖=γ€–π‘π‘œπ‘  〗⁑〖(πœ‹/2 βˆ’π‘₯)γ€— γ€— ) ("As " (𝑑(π‘₯))/𝑑π‘₯ " = 1 & " πœ‹/2 " is constant" ) Example 45 (Method 2) Differentiate the following 𝑀.π‘Ÿ.𝑑. π‘₯. (i) cos^(βˆ’1) (sin⁑π‘₯) Let 𝑓(π‘₯) = cos^(βˆ’1) (sin⁑π‘₯) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑓′(π‘₯) = (βˆ’1)/√(1 βˆ’ γ€–(sin⁑π‘₯)γ€—^2 ) Γ— (sin⁑π‘₯ )^β€² 𝑓′(π‘₯) = (βˆ’1)/√(1 βˆ’ sin^2⁑π‘₯ ) Γ—cos⁑π‘₯ 𝑓′(π‘₯) = (βˆ’1)/√(cos^2⁑π‘₯ ) Γ—cos⁑π‘₯ 𝑓′(π‘₯) = (βˆ’1)/cos⁑π‘₯ Γ—cos⁑π‘₯ 𝒇’(𝒙) = βˆ’1 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 Example 45 Differentiate the following w.r.t. x. (iii) sin^(βˆ’1) ((2^( π‘₯+1) )/( 1 +γ€– 4 γ€—^π‘₯ )) Let 𝑓(π‘₯) = sin^(βˆ’1) ((2^( π‘₯+1) )/( 1 +γ€– 4 γ€—^π‘₯ )) 𝑓(π‘₯) = sin^(βˆ’1) ((2^( π‘₯). 2)/( 1 + (2^π‘₯ )^2 )) Let 𝟐^𝒙 = tan ΞΈ 𝑓(π‘₯) = sin^(βˆ’1) ((tanβ‘γ€–πœƒ γ€—. 2)/( 1 + tan^2β‘πœƒ )) = sin^(βˆ’1) ((2 tanβ‘γ€–πœƒ γ€— )/( 1 +γ€– tan^2γ€—β‘πœƒ )) = sin^(βˆ’1) (sin 2πœƒ) = 2πœƒ (sin⁑2πœƒ "= " (2 tanβ‘πœƒ)/(1 +γ€– tan^2γ€—β‘πœƒ )) (As 〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖(π‘ π‘–π‘›β‘πœƒ)γ€— =πœƒ) Since 2^π‘₯= tanβ‘πœƒ tan^(βˆ’1) (2^π‘₯ )=πœƒ ∴ 𝒇(𝒙) = 𝟐 (〖𝒕𝒂𝒏〗^(βˆ’πŸ) (𝟐^𝒙 )) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑓’(π‘₯) = 2 (𝑑 (tan^(βˆ’1) 2^π‘₯ )" " )/𝑑π‘₯ 𝑓’(π‘₯) = 2 . 1/(1 + (2^π‘₯ )^2 ) . (𝒅 (𝟐^𝒙 )" " )/𝒅𝒙 𝑓’(π‘₯) = (2 )/(1 + (2^π‘₯ )^2 ) . 𝟐^𝒙 . π’π’π’ˆβ‘πŸ (𝐴𝑠 𝑑/𝑑π‘₯(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1))=1/(1+π‘₯^2 )) (𝐴𝑠 𝑑/𝑑π‘₯ (π‘Ž^π‘₯ )=π‘Ž^π‘₯. π‘™π‘œπ‘”β‘π‘₯ ) 𝑓’(π‘₯) = (γ€–2. 2γ€—^π‘₯.γ€– log〗⁑2)/(1 + (2^π‘₯ )^2 ) 𝑓’(π‘₯) = (2^(π‘₯ + 1).γ€– log〗⁑2)/(1 + (2^π‘₯ )^2 ) 𝑓’(π‘₯) = (2^(π‘₯ + 1).γ€– log〗⁑2)/(1 + (2^2 )^π‘₯ ) 𝒇’(𝒙) = (𝟐^(𝒙 + 𝟏).γ€– π’π’π’ˆγ€—β‘πŸ)/(𝟏 + πŸ’^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.