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Misc  5 - Differentiate cos-1 x/2 / root (2x + 7) - Teachoo

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

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Misc 5 Differentiate 𝑤.𝑟.𝑡. 𝑥 the function, (〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2)/√(2𝑥+7 ) , – 2 < 𝑥 < 2 Let 𝑦= (〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2)/√(2𝑥+7 ) Differentiating both sides 𝑤.𝑟.𝑡. 𝑥 𝑑𝑦/𝑑𝑥 = 𝑑/𝑑𝑥 ((〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2)/√(2𝑥+7 )) Using Quotient rule As (𝑢/𝑣)^′ = (𝑢^′ 𝑣 − 𝑣^′ 𝑢)/𝑣^2 where u = cos−1 𝑥/2 & v = √(2𝑥+7) 𝑑𝑦/𝑑𝑥 = (𝑑(〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2)/𝑑𝑥 . √(2𝑥 + 7 ) − 𝑑(√(2𝑥 + 7 ))/𝑑𝑥 . 〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2 )/(√(2𝑥 + 7 ))^2 𝑑𝑦/𝑑𝑥 = ((−1)/√(1 −(𝑥/2)^2 ) . 𝑑(𝑥/2)/𝑑𝑥 √(2𝑥 + 7 ) − 1/(2√(2𝑥 + 7)) . 𝑑(2𝑥 + 7)/𝑑𝑥 . 〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2 )/(√(2𝑥 + 7 ))^2 𝑑𝑦/𝑑𝑥 = ((−1)/√(〖(4 − 𝑥)/4〗^2 ) . 1/2 √(2𝑥 + 7 ) − 1/(2√(2𝑥 + 7)) . (2 + 0) . 〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2 )/((2𝑥 + 7) ) 𝑑𝑦/𝑑𝑥 = ((−2)/√(〖4 − 𝑥〗^2 ) . 1/2 √(2𝑥 + 7 ) − 1/(2√(2𝑥 + 7)) . 2 . 〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2 )/((2𝑥 + 7) ) 𝑑𝑦/𝑑𝑥 = ((− √(𝟐𝒙 + 𝟕 ))/( √(〖𝟒 − 𝒙〗^𝟐 )) − (. 〖𝒄𝒐𝒔〗^(−𝟏 ) 𝒙/𝟐)/√(𝟐𝒙 + 𝟕) )/((2𝑥 + 7) ) 𝑑𝑦/𝑑𝑥 = (− √(2𝑥 + 7 ) ( √(2𝑥 + 7 )) − 〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2 √(〖4 − 𝑥〗^2 ))/((2𝑥 + 7) (√(2𝑥 + 7 )) (√(〖4 − 𝑥〗^2 )) ) 𝑑𝑦/𝑑𝑥 = 〖−(√(2𝑥 + 7 ))〗^2/((2𝑥 + 7) ( √(2𝑥 + 7 )) (√(〖4 − 𝑥〗^2 )) ) "+ " (〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2 √(〖4 − 𝑥〗^2 ))/((2𝑥 + 7) ( √(2𝑥 + 7 )) (√(〖4 − 𝑥〗^2 )) ) 𝑑𝑦/𝑑𝑥 = (−(2𝑥 + 7))/((2𝑥 + 7) √(2𝑥 + 7 ) √(〖4 − 𝑥〗^2 )) "+ " (〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2 )/((2𝑥 + 7) √(2𝑥 + 7 ) ) 𝑑𝑦/𝑑𝑥 = (−1)/(√(2𝑥 + 7 ) √(〖4 − 𝑥〗^2 )) "+ " (〖𝑐𝑜𝑠〗^(−1 ) 𝑥/2 )/((2𝑥 + 7) (2𝑥 + 7)^(1/2) ) 𝒅𝒚/𝒅𝒙 = (−𝟏)/(√(𝟐𝒙 + 𝟕) √(〖𝟒 − 𝒙〗^𝟐 ))−(〖𝒄𝒐𝒔〗^(−𝟏 ) 𝒙/𝟐 )/((𝟐𝒙 + 𝟕)^(𝟑/𝟐) )

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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.