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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.5, 16 Find the derivative of the function given by f (๐‘ฅ) = (1 + ๐‘ฅ) (1 + ๐‘ฅ^2) (1 + ๐‘ฅ^4) (1 + ๐‘ฅ8) and hence find f โ€ฒ(1) . Given ๐‘“(๐‘ฅ)=(1+๐‘ฅ)(1+๐‘ฅ^2 )(1+๐‘ฅ^4 )(1+๐‘ฅ^8 )" " Let ๐‘ฆ=(1+๐‘ฅ)(1+๐‘ฅ^2 )(1+๐‘ฅ^4 )(1+๐‘ฅ^8 ) Taking log both sides log ๐‘ฆ = log (1+๐‘ฅ)(1+๐‘ฅ^2 )(1+๐‘ฅ^4 )(1+๐‘ฅ^8 ) log ๐‘ฆ = log (1+๐‘ฅ)+logโก(1+๐‘ฅ^2 )+logโก(1+๐‘ฅ^4 ) ใ€–+ logใ€—โกใ€– (1+๐‘ฅ^8 )ใ€— (As ๐‘™๐‘œ๐‘”โก(๐‘Ž๐‘)= ๐‘™๐‘œ๐‘”โก๐‘Ž + ๐‘™๐‘œ๐‘”โก๐‘) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘(logโก๐‘ฆ )/๐‘‘๐‘ฅ = ๐‘‘(log (1 + ๐‘ฅ) + logโก(1 + ๐‘ฅ^2 ) + logโก(1 + ๐‘ฅ^4 )+ logโกใ€– (1 + ๐‘ฅ^8 )ใ€— )/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ฆ )/๐‘‘๐‘ฅ = ๐‘‘(log (1 + ๐‘ฅ))/๐‘‘๐‘ฅ + ๐‘‘(logโก(1 + ๐‘ฅ^2 ) )/๐‘‘๐‘ฅ + ๐‘‘(logโก(1 + ๐‘ฅ^4 ) )/๐‘‘๐‘ฅ + ๐‘‘(logโกใ€– (1 + ๐‘ฅ^8 )ใ€— )/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ฆ )/๐‘‘๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/(1 + ๐‘ฅ) . ๐‘‘(1 + ๐‘ฅ)/๐‘‘๐‘ฅ + 1/((1 + ๐‘ฅ^2 ) ) . ๐‘‘(1 + ๐‘ฅ^2 )/๐‘‘๐‘ฅ + 1/((1 + ๐‘ฅ^4 ) ) . ๐‘‘(1 + ๐‘ฅ^4 )/๐‘‘๐‘ฅ + 1/((1 + ๐‘ฅ^8 ) ) . ๐‘‘(1 + ๐‘ฅ^8 )/๐‘‘๐‘ฅ 1/๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/(1 + ๐‘ฅ) . (0+1) + 1/((1 + ๐‘ฅ^2 ) ) . (0+2๐‘ฅ) + 1/((1 + ๐‘ฅ^4 ) ) . (0+4๐‘ฅ^3 ) + 1/((1 + ๐‘ฅ^8 ) ) . (0+8๐‘ฅ^7 ) 1/๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/(1 + ๐‘ฅ) + 2๐‘ฅ/(1 + ๐‘ฅ^2 ) + (4๐‘ฅ^3)/(1 + ๐‘ฅ^4 ) + (8๐‘ฅ^7)/(1 + ๐‘ฅ^8 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฆ (1/(1 + ๐‘ฅ) " + " 2๐‘ฅ/(1 + ๐‘ฅ^2 ) " + " (4๐‘ฅ^3)/(1 + ๐‘ฅ^4 ) " + " (8๐‘ฅ^7)/(1 + ๐‘ฅ^8 )) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (1+๐‘ฅ)(1+๐‘ฅ^2 )(1+๐‘ฅ^4 )(1+๐‘ฅ^8 ) (1/(1 + ๐‘ฅ) " + " 2๐‘ฅ/(1 + ๐‘ฅ^2 ) " + " (4๐‘ฅ^3)/(1 + ๐‘ฅ^4 ) " +" (8๐‘ฅ^7)/(1 + ๐‘ฅ^8 )) Hence, ๐’‡โ€ฒ(๐’™) = (๐Ÿ+๐’™)(๐Ÿ+๐’™^๐Ÿ )(๐Ÿ+๐’™^๐Ÿ’ )(๐Ÿ+๐’™^๐Ÿ– ) (๐Ÿ/(๐Ÿ + ๐’™) " + " ๐Ÿ๐’™/(๐Ÿ + ๐’™^๐Ÿ ) " + " (๐Ÿ’๐’™^๐Ÿ‘)/(๐Ÿ + ๐’™^๐Ÿ’ ) " + " (๐Ÿ–๐’™^๐Ÿ•)/(๐Ÿ + ๐’™^๐Ÿ– )) We need to find ๐‘“โ€ฒ(1) Putting ๐‘ฅ=1 ๐‘“โ€ฒ(1) = (1+1)(1+(1)^2 )(1+(1)^4 )(1+ใ€–(1)ใ€—^8 ) (1/(1 +1) " + " 2(1)/(1+(1)^2 ) " + " (4(1)^3)/(1 + (1)^4 ) " + " (8(1)^7)/(1 + (1)^8 )) = 2(1+1)(1+1)(1+1) (1/(1 + 1) " + " 2/(1 + 1) " + " 4/(1 + 1) " + " 8/(1 + 1)) = 2(2)(2)(2) (1/2 " + " 2/2 " + " 4/2 " + " 8/2) = 16 ((1 + 2 + 4 + 8)/2) = 16 ร— 15/2 = 8 "ร— " 15 = 120

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.