Ex 5.5, 7 - Differentiate (log x)x + x log x - Chapter 5 - Ex 5.5

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.5, 7 (Method 1) Differentiate the functions in, ใ€–(logโกใ€–๐‘ฅ)ใ€—ใ€—^๐‘ฅ + ๐‘ฅ^logโก๐‘ฅ Let ๐‘ฆ = ใ€–(logโกใ€–๐‘ฅ)ใ€—ใ€—^๐‘ฅ+ ๐‘ฅ^logโก๐‘ฅ Let ๐‘ข = ใ€–(logโกใ€–๐‘ฅ)ใ€—ใ€—^๐‘ฅ , ๐‘ฃ = ๐‘ฅ^logโก๐‘ฅ ๐‘ฆ = ๐‘ข+๐‘ฃ Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘ (๐‘ข + ๐‘ฃ))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ข/๐‘‘๐‘ฅ + ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" 1 . log (log" " ๐‘ฅ) + (1/(log" " ๐‘ฅ) .๐‘‘(log" " ๐‘ฅ)/๐‘‘๐‘ฅ) ร— ๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (log" " ๐‘ฅ) + (1/(log" " ๐‘ฅ) . 1/๐‘ฅ) ร— ๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (log" " ๐‘ฅ) + 1/logโก๐‘ฅ ร— ๐‘ฅ/๐‘ฅ 1/๐‘ข (๐‘‘๐‘ข/๐‘‘๐‘ฅ)" =" log (log" " ๐‘ฅ) + 1/logโก๐‘ฅ ๐‘‘๐‘ข/๐‘‘๐‘ฅ " =" ๐‘ข (log (log" " ๐‘ฅ)" + " 1/logโก๐‘ฅ ) ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (logโก๐‘ฅ )^๐‘ฅ (log (log" " ๐‘ฅ)" + " 1/logโก๐‘ฅ ) ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (logโก๐‘ฅ )^๐‘ฅ ((logโกใ€–๐‘ฅ . ใ€–log ใ€—โก(logโก๐‘ฅ ) +ใ€— 1)/logโก๐‘ฅ ) ๐‘‘๐‘ข/๐‘‘๐‘ฅ = (logโก๐‘ฅ )^๐‘ฅ/logโก๐‘ฅ (logโกใ€–๐‘ฅ . ใ€–log ใ€—โก(logโก๐‘ฅ )+ใ€— 1)

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