Ex 5.6, 1 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at April 16, 2024 by

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Ex 5.6, 1 If x and y are connected parametrically by the equations without eliminating the parameter, Find ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ, x = 2ใ€–๐‘Ž๐‘กใ€—^2, y = ใ€–๐‘Ž๐‘กใ€—^4Here ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘๐‘ฆ/๐‘‘๐‘ก)/(๐‘‘๐‘ฅ/๐‘‘๐‘ก) Calculating ๐’…๐’š/๐’…๐’• ๐‘ฆ = ใ€–๐‘Ž๐‘กใ€—^4 ๐‘‘๐‘ฆ/๐‘‘๐‘ก " =" ใ€–4๐‘Ž๐‘กใ€—^(4 โˆ’1) "=" ใ€–4๐‘Ž๐‘กใ€—^3 Calculating ๐’…๐’™/๐’…๐’• ๐‘ฅ = 2ใ€–๐‘Ž๐‘กใ€—^2 ๐‘‘๐‘ฅ/๐‘‘๐‘ก " =" 2 ร— 2๐‘Ž๐‘ก "=" 4๐‘Ž๐‘ก Now, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘๐‘ฆ/๐‘‘๐‘ก)/(๐‘‘๐‘ฅ/๐‘‘๐‘ก) Putting values ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (4๐‘Ž๐‘ก^3)/4๐‘Ž๐‘ก ๐’…๐’š/๐’…๐’™ = t2

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