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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 40 (Method 1) 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 40 (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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.