Misc 5 - Chapter 5 Class 12 Continuity and Differentiability

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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/𝑑𝑥 (𝑥/2) √(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 = (−(2𝑥 + 7))/((2𝑥 + 7) √(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) )