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Ex 5.5, 16 - Find derivative of f(x)=(1+x)(1+x2)(1+x4)(1+x8)

Ex 5.5, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.5, 16 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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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€— 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) 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)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.