Check sibling questions

Misc 6 - Differentiate cot^-1 [ root (1+sinx) + root (1 - sin x)]

Misc  6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc  6 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Misc  6 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Misc  6 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Misc  6 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Misc  6 - Chapter 5 Class 12 Continuity and Differentiability - Part 7

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

Misc 6 (Method 1) Differentiate w.r.t. x the function, γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ ))/(√(1 + sin⁑〖π‘₯ γ€— ) βˆ’ √(1 βˆ’sin⁑π‘₯ )) ] , 0<π‘₯< πœ‹/2 Let 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(√(1 + sin⁑π‘₯ ) +√(1 βˆ’ sin⁑π‘₯ ))/(√(1 + sin⁑〖π‘₯ γ€— )βˆ’βˆš(1 βˆ’ sin⁑π‘₯ )) ] Rationalizing the sum 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [((√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ )))/((√(1 + sin⁑〖π‘₯ γ€— )βˆ’ √(1 βˆ’ sin⁑π‘₯ )) ) Γ—((√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ )))/((√(1+sin⁑〖π‘₯ γ€— )+ √(1 βˆ’ sin⁑π‘₯ )) )] 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ ))^2/((√(1 + sin⁑〖π‘₯ γ€— )βˆ’ √(1 βˆ’γ€– sin〗⁑π‘₯ )) (√(1 + sin⁑〖π‘₯ γ€— )+ √(1 βˆ’γ€– sin〗⁑π‘₯ )) ) ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [((√(1 + sin⁑π‘₯ ) )^2 + (√(1 βˆ’ sin⁑π‘₯ ) )^2+ 2(√(1 + sin⁑〖π‘₯ γ€— ))(√(1 βˆ’γ€– sin〗⁑π‘₯ )))/((√(1 + sin⁑〖π‘₯ γ€— )βˆ’ √(1 βˆ’ γ€– sin〗⁑π‘₯ )) (√(1 + sin⁑〖π‘₯ γ€— )+ √(1 βˆ’ sin⁑π‘₯ )) ) ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [((1 + sin⁑π‘₯ ) + (1 βˆ’ sin⁑π‘₯ ) + 2√((1 + sin⁑π‘₯ ) (1 βˆ’ sin⁑π‘₯ ) ))/((√(1 + sin⁑〖π‘₯ γ€— ))^2 βˆ’ (√(1 βˆ’ sin⁑π‘₯ ))^2 ) ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(1 + sin⁑π‘₯ + 1 βˆ’ sin⁑π‘₯ + 2√((1)^2 βˆ’ sin^2⁑π‘₯ ))/(1 + sin⁑π‘₯ βˆ’ 1 + sin⁑π‘₯ ) ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(2 + 2√(1 βˆ’ sin^2⁑π‘₯ ))/(2 sin⁑π‘₯ ) ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(2 (1 + √(𝟏 βˆ’ γ€–π’”π’Šπ’γ€—^πŸβ‘π’™ ) ) )/(2 sin⁑π‘₯ )] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(1 + √(〖𝒄𝒐𝒔〗^πŸβ‘π’™ ) )/sin⁑π‘₯ ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(1 + 𝒄𝒐𝒔⁑𝒙 )/π’”π’Šπ’β‘π’™ ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(1 + 𝟐 〖𝒄𝒐𝒔〗^πŸβ‘γ€–π’™/πŸγ€— βˆ’ 𝟏 )/(𝟐 π’”π’Šπ’β‘γ€– 𝒙/𝟐 γ€— γ€–πœπ¨π¬ 〗⁑〖𝒙/πŸγ€— )] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(2 cos^2⁑〖π‘₯/2γ€— )/(2 sin⁑〖 π‘₯/2 γ€— γ€–cos 〗⁑〖π‘₯/2γ€— )] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(γ€–cos 〗⁑〖π‘₯/2γ€— )/sin⁑〖 π‘₯/2 γ€— ] = γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [cot⁑(π‘₯/2) ] = π‘₯/2 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 Thus, π’š= 𝒙/𝟐 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑/𝑑π‘₯ (π‘₯/2) 𝑑𝑦/𝑑π‘₯ = 1/2 𝑑π‘₯/𝑑π‘₯ π’…π’š/𝒅𝒙 = 𝟏/𝟐 Misc 6 (Method 2) Differentiate w.r.t. x the function, γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ ))/(√(1 + sin⁑〖π‘₯ γ€— ) βˆ’ √(1 βˆ’ sin⁑π‘₯ )) ] , 0<π‘₯< πœ‹/2 Let 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ ))/(√(1 +γ€– sin〗⁑〖π‘₯ γ€— ) βˆ’ √(1 βˆ’ sin⁑π‘₯ )) ] Finding √(𝟏 + π’”π’Šπ’β‘π’™ ) and √(𝟏 βˆ’ π’”π’Šπ’β‘π’™ ) γ€–π’”π’Šπ’γ€—^𝟐 𝜽+〖𝒄𝒐𝒔〗^𝟐⁑𝜽=1 Replacing πœƒ by π‘₯/2 𝑠𝑖𝑛2 π‘₯/2 + γ€–π‘π‘œπ‘ γ€—^2 π‘₯/2 = 1 π’”π’Šπ’β‘πŸπœ½=2 π‘ π‘–π‘›β‘γ€–πœƒ π‘π‘œπ‘ β‘πœƒ γ€— Replacing πœƒ by π‘₯/2 π’”π’Šπ’β‘π’™ = 2 𝑠𝑖𝑛⁑π‘₯/2 π‘π‘œπ‘ β‘π‘₯/2 √("1 + sin x" ) = √((〖𝑠𝑖𝑛〗^2 π‘₯/2 + γ€–π‘π‘œπ‘ γ€—^2 π‘₯/2)" + 2 sin " π‘₯/2 " cos " π‘₯/2) = √((π‘π‘œπ‘  π‘₯/2 +sin⁑〖π‘₯/2γ€— )^2 ) = 𝒄𝒐𝒔 𝒙/𝟐 +π’”π’Šπ’β‘γ€–π’™/πŸγ€— √("1 " βˆ’" sin x" ) = √((〖𝑠𝑖𝑛〗^2 π‘₯/2 + γ€–π‘π‘œπ‘ γ€—^2 π‘₯/2)" βˆ’ 2 sin " π‘₯/2 " cos " π‘₯/2) = √((π‘π‘œπ‘  π‘₯/2 βˆ’sin⁑〖π‘₯/2γ€— )^2 ) = 𝒄𝒐𝒔 𝒙/𝟐 +π’”π’Šπ’β‘γ€–π’™/πŸγ€— √("1 " βˆ’" sin x" ) = √((〖𝑠𝑖𝑛〗^2 π‘₯/2 + γ€–π‘π‘œπ‘ γ€—^2 π‘₯/2)" βˆ’ 2 sin " π‘₯/2 " cos " π‘₯/2) = √((π‘π‘œπ‘  π‘₯/2 βˆ’sin⁑〖π‘₯/2γ€— )^2 ) = 𝒄𝒐𝒔 𝒙/𝟐 +π’”π’Šπ’β‘γ€–π’™/πŸγ€— Thus, our equation becomes y = γ€–π‘π‘œπ‘ γ€—^(βˆ’1) |(√(1 + sin⁑π‘₯ ) + √(1 βˆ’ sin⁑π‘₯ ))/(√(1 + sin⁑π‘₯ ) βˆ’ √(1 βˆ’ sin⁑π‘₯ ))| Substituting value of √(1+𝑠𝑖𝑛⁑π‘₯ ) & √(1βˆ’π‘ π‘–π‘›β‘π‘₯ ) from (1) & (2). y = cotβˆ’1 [((γ€–cos 〗⁑〖π‘₯/2γ€— + γ€–sin 〗⁑〖π‘₯/2γ€— ) + (γ€–cos 〗⁑〖π‘₯/2γ€— βˆ’ γ€–sin 〗⁑〖π‘₯/2γ€— ))/((γ€–cos 〗⁑〖π‘₯/2γ€— + γ€–sin 〗⁑〖π‘₯/2γ€— ) + (γ€–cos 〗⁑〖π‘₯/2γ€— βˆ’ γ€–sin 〗⁑〖π‘₯/2γ€— ) )] 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(π‘π‘œπ‘ β‘γ€– π‘₯/2γ€— + 〖𝑠𝑖𝑛 〗⁑〖π‘₯/2γ€— + π‘π‘œπ‘ β‘γ€– π‘₯/2γ€— βˆ’ 〖𝑠𝑖𝑛 〗⁑〖π‘₯/2γ€—)/(π‘π‘œπ‘ β‘γ€– π‘₯/2γ€— + 𝑠𝑖𝑛⁑〖 π‘₯/2γ€— βˆ’ γ€–π‘π‘œπ‘  〗⁑〖π‘₯/2γ€— βˆ’ 𝑠𝑖𝑛⁑〖 π‘₯/2γ€— ) ] 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [(2 γ€–π‘π‘œπ‘  〗⁑〖π‘₯/2γ€— )/(2 𝑠𝑖𝑛⁑〖 π‘₯/2γ€— ) ] 𝑦= γ€–π‘π‘œπ‘‘γ€—^(βˆ’1 ) [π‘π‘œπ‘‘ (π‘₯ )/2 ] π’š= 𝒙/𝟐 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(π‘₯/2)/𝑑π‘₯ π’…π’š/𝒅𝒙 = 𝟏/𝟐

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.