If y = log (√x + 1 / √x) then prove that x (x + 1) 2 y 2 + (x + 1) 2 y 1 = 2

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Question 15 If y = log (โˆš๐‘ฅ+1/โˆš๐‘ฅ)^2then prove that x (x + 1)2 y2 + (x + 1)2 y1 = 2 Given y = log (โˆš๐‘ฅ+1/โˆš๐‘ฅ)^2 Using log a2 = 2 log a y = 2 log (โˆš๐‘ฅ+1/โˆš๐‘ฅ) y = 2 log ((โˆš๐‘ฅ ร— โˆš๐‘ฅ + 1)/โˆš๐‘ฅ) y = 2 log ((๐‘ฅ + 1)/โˆš๐‘ฅ) Using log ๐‘Ž/๐‘ = log a โ€“ log b y = 2 log (๐‘ฅ+1) โ€“ 2 log โˆš๐‘ฅ y = 2 log (๐‘ฅ+1) โ€“ 2 log (๐‘ฅ)^(1/2) Using log a2 = 2 log a y = 2 log (๐‘ฅ+1) โ€“ 2 ร— 1/2 log ๐‘ฅ y = 2 log (๐‘ฅ+1) โ€“ log ๐‘ฅ Now finding y1 and y2 From (1) y = 2 log (๐‘ฅ+1) โ€“ log ๐‘ฅ Differentiating w.r.t x ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2 ๐‘‘(logโก(๐‘ฅ + 1) )/๐‘‘๐‘ฅ โ€“ ๐‘‘(logโก๐‘ฅ )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2 1/(๐‘ฅ + 1) โ€“ 1/๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2/(๐‘ฅ + 1) โ€“ 1/๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (2๐‘ฅ โˆ’(๐‘ฅ +1))/(๐‘ฅ(๐‘ฅ + 1)) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((๐‘ฅ โˆ’ 1))/(๐‘ฅ(๐‘ฅ + 1)) y1 = ((๐‘ฅ โˆ’ 1))/(๐‘ฅ(๐‘ฅ + 1)) Now, from (2) y1 = ((๐‘ฅ โˆ’ 1))/(๐‘ฅ(๐‘ฅ + 1)) Differentiating w.r.t x y2 = (((๐‘ฅ โˆ’ 1))/๐‘ฅ(๐‘ฅ + 1) )^โ€ฒ = x (x + 1)2 ร— (ใ€–โˆ’๐‘ฅใ€—^2 +2๐‘ฅ + 1)/(๐‘ฅ^2 (๐‘ฅ + 1)^2 ) + (x + 1)2 ร— ((๐‘ฅ โˆ’ 1))/(๐‘ฅ(๐‘ฅ + 1)) = (ใ€–โˆ’๐‘ฅใ€—^2 +2๐‘ฅ + 1)/๐‘ฅ + (x + 1) ร— ((๐‘ฅ โˆ’ 1))/๐‘ฅ = (ใ€–โˆ’๐‘ฅใ€—^2 +2๐‘ฅ + 1)/๐‘ฅ + ((๐‘ฅ + 1)(๐‘ฅ โˆ’ 1))/๐‘ฅ = (ใ€–โˆ’๐‘ฅใ€—^2 +2๐‘ฅ + 1)/๐‘ฅ + (๐‘ฅ^2 โˆ’ 1)/๐‘ฅ = (ใ€–โˆ’๐‘ฅใ€—^2 +2๐‘ฅ + 1 + ๐‘ฅ^2 โˆ’ 1)/๐‘ฅ = 2๐‘ฅ/๐‘ฅ = 2 = RHS Hence proved

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