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Ex 5.5

Ex 5.5, 1
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Last updated at March 16, 2023 by Teachoo

Ex 5.5, 3 Differentiate the functions in, (log𝑥 )^cos𝑥 Let 𝑦=(log𝑥 )^cos𝑥 Taking log both sides log𝑦 = log〖〖 (log𝑥 )〗^cos𝑥 〗 log𝑦 = cos〖𝑥 .〖log 〗(log𝑥 ) 〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑(log𝑦)/𝑑𝑥 = 𝑑(cos〖𝑥 .〖 log 〗(log𝑥 ) 〗 )/𝑑𝑥 𝑑(log𝑦 )/𝑑𝑥 (𝑑𝑦/𝑑𝑦) = 𝑑(cos〖𝑥 .〖 log 〗(log𝑥 ) 〗 )/𝑑𝑥 𝑑(log𝑦 )/𝑑𝑥 (𝑑𝑦/𝑑𝑥) = 𝑑(cos〖𝑥 .〖 log 〗(log𝑥 ) 〗 )/𝑑𝑥 (As 𝑙𝑜𝑔(𝑎^𝑏) = 𝑏 𝑙𝑜𝑔𝑎) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 𝑑(cos〖𝑥 .〖 log 〗(log𝑥 ) 〗 )/𝑑𝑥 1/𝑦 𝑑𝑦/𝑑𝑥 = 𝑑(cos𝑥 )/𝑑𝑥 . 〖 log 〗(log𝑥 ) + 𝑑(〖 log 〗(log𝑥 ) )/𝑑𝑥 . cos𝑥 1/𝑦 𝑑𝑦/𝑑𝑥 = 〖−sin〗𝑥 . 〖log 〗(log𝑥 ) + 1/log𝑥 . 𝑑(log𝑥 )/𝑑𝑥 . cos𝑥 1/𝑦 𝑑𝑦/𝑑𝑥 = 〖−sin〗𝑥 . 〖log 〗(log𝑥 ) + 1/log𝑥 × 1/𝑥 . cos𝑥 1/𝑦 𝑑𝑦/𝑑𝑥 = 〖−sin〗𝑥 . 〖log 〗(log𝑥 ) + cos𝑥/(𝑥 log𝑥 ) 𝑑𝑦/𝑑𝑥 = 𝑦 (〖−sin〗𝑥 " . " 〖log 〗(log𝑥 )" + " cos𝑥/(𝑥 log𝑥 )) Using product rule in 𝑐𝑜𝑠〖𝑥 .〖 𝑙𝑜𝑔 〗(𝑙𝑜𝑔𝑥 ) 〗 (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 Putting values of 𝑦 𝑑𝑦/𝑑𝑥 = (𝑙𝑜𝑔𝑥 )^𝑐𝑜𝑠𝑥 (〖−𝑠𝑖𝑛〗𝑥 " . " 〖𝑙𝑜𝑔 〗(𝑙𝑜𝑔𝑥 )" + " 𝑐𝑜𝑠𝑥/(𝑥 𝑙𝑜𝑔𝑥 )) 𝒅𝒚/𝒅𝒙 = (𝒍𝒐𝒈𝒙 )^𝒄𝒐𝒔𝒙 (𝒄𝒐𝒔𝒙/(𝒙 𝒍𝒐𝒈𝒙 ) 〖 − 𝒔𝒊𝒏〗𝒙 " . " 〖𝐥𝐨𝐠 〗(𝒍𝒐𝒈𝒙 ) )