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


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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

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.