Misc 4 - Differentiate sin-1 (x root x) - Chapter 5 NCERT - Miscellaneous

part 2 - Misc  4 - Miscellaneous - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability

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Transcript

Misc 4 Differentiate ๐‘ค.๐‘Ÿ.๐‘ก. ๐‘ฅ the function, ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (๐‘ฅ โˆš๐‘ฅ), 0 โ‰ค ๐‘ฅ โ‰ค 1 Let ๐‘ฆ=ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (๐‘ฅ โˆš๐‘ฅ) ๐‘ฆ=ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (๐‘ฅ . ๐‘ฅ^(1/2)) ๐‘ฆ=ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (๐‘ฅ^(1 + 1/2)) ๐’š=ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ) (๐’™^( ๐Ÿ‘/๐Ÿ) ) Differentiating ๐‘ค.๐‘Ÿ.๐‘ก. ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (๐‘ฅ^( 3/2)))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/โˆš(1 โˆ’ (๐‘ฅ^(3/2) )^2 ) ร— (๐‘‘(๐‘ฅ)^(3/2))/๐‘‘๐‘ฅ ("As " ๐‘‘(ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โก๐‘ฅ )/๐‘‘๐‘ฅ=1/โˆš(1 โˆ’ ๐‘ฅ^2 )) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/โˆš(1 โˆ’ ๐‘ฅ^3 ) ร— 3/2 (๐‘ฅ)^(3/2 โˆ’1) = 1/โˆš(1 โˆ’ ๐‘ฅ^3 ) ร— 3/2 ใ€–๐‘ฅ ใ€—^(1/2 ) = 1/โˆš(1 โˆ’ ๐‘ฅ^3 ) ร— 3/2 โˆš๐‘ฅ = ๐Ÿ‘/๐Ÿ โˆš(๐’™/(๐Ÿ โˆ’๐’™^๐Ÿ‘ ))

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