# Ex 5.5, 16 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at April 16, 2024 by Teachoo

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Ex 5.5, 16 Important You are here

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Last updated at April 16, 2024 by Teachoo

Ex 5.5, 16 Find the derivative of the function given by f (๐ฅ) = (1 + ๐ฅ) (1 + ๐ฅ^2) (1 + ๐ฅ^4) (1 + ๐ฅ8) and hence find f โฒ(1) .Given ๐(๐ฅ)=(1+๐ฅ)(1+๐ฅ^2 )(1+๐ฅ^4 )(1+๐ฅ^8 )" " Let ๐ฆ=(1+๐ฅ)(1+๐ฅ^2 )(1+๐ฅ^4 )(1+๐ฅ^8 ) Taking log both sides log ๐ฆ = log (1+๐ฅ)(1+๐ฅ^2 )(1+๐ฅ^4 )(1+๐ฅ^8 ) log ๐ฆ = log (1+๐ฅ)+logโก(1+๐ฅ^2 )+logโก(1+๐ฅ^4 ) ใ+ logใโกใ (1+๐ฅ^8 )ใ Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐(logโก๐ฆ )/๐๐ฅ = ๐(log (1 + ๐ฅ) + logโก(1 + ๐ฅ^2 ) + logโก(1 + ๐ฅ^4 )+ logโกใ (1 + ๐ฅ^8 )ใ )/๐๐ฅ ๐(logโก๐ฆ )/๐๐ฅ = ๐(log (1 + ๐ฅ))/๐๐ฅ + ๐(logโก(1 + ๐ฅ^2 ) )/๐๐ฅ + ๐(logโก(1 + ๐ฅ^4 ) )/๐๐ฅ + ๐(logโกใ (1 + ๐ฅ^8 )ใ )/๐๐ฅ ๐(logโก๐ฆ )/๐๐ฆ . ๐๐ฆ/๐๐ฅ = 1/(1 + ๐ฅ) . ๐(1 + ๐ฅ)/๐๐ฅ + 1/((1 + ๐ฅ^2 ) ) . ๐(1 + ๐ฅ^2 )/๐๐ฅ + 1/((1 + ๐ฅ^4 ) ) . ๐(1 + ๐ฅ^4 )/๐๐ฅ + 1/((1 + ๐ฅ^8 ) ) . ๐(1 + ๐ฅ^8 )/๐๐ฅ 1/๐ฆ . ๐๐ฆ/๐๐ฅ = 1/(1 + ๐ฅ) . (0+1) + 1/((1 + ๐ฅ^2 ) ) . (0+2๐ฅ) + 1/((1 + ๐ฅ^4 ) ) . (0+4๐ฅ^3 ) + 1/((1 + ๐ฅ^8 ) ) . (0+8๐ฅ^7 ) 1/๐ฆ . ๐๐ฆ/๐๐ฅ = 1/(1 + ๐ฅ) + 2๐ฅ/(1 + ๐ฅ^2 ) + (4๐ฅ^3)/(1 + ๐ฅ^4 ) + (8๐ฅ^7)/(1 + ๐ฅ^8 ) ๐๐ฆ/๐๐ฅ = ๐ฆ (1/(1 + ๐ฅ) " + " 2๐ฅ/(1 + ๐ฅ^2 ) " + " (4๐ฅ^3)/(1 + ๐ฅ^4 ) " + " (8๐ฅ^7)/(1 + ๐ฅ^8 )) ๐๐ฆ/๐๐ฅ = (1+๐ฅ)(1+๐ฅ^2 )(1+๐ฅ^4 )(1+๐ฅ^8 ) (1/(1 + ๐ฅ) " + " 2๐ฅ/(1 + ๐ฅ^2 ) " + " (4๐ฅ^3)/(1 + ๐ฅ^4 ) " +" (8๐ฅ^7)/(1 + ๐ฅ^8 )) Hence, ๐โฒ(๐) = (๐+๐)(๐+๐^๐ )(๐+๐^๐ )(๐+๐^๐ ) (๐/(๐ + ๐) " + " ๐๐/(๐ + ๐^๐ ) " + " (๐๐^๐)/(๐ + ๐^๐ ) " + " (๐๐^๐)/(๐ + ๐^๐ )) We need to find ๐โฒ(1) Putting ๐ฅ=1 ๐โฒ(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) ๐๐ฆ/๐๐ฅ = (1+๐ฅ)(1+๐ฅ^2 )(1+๐ฅ^4 )(1+๐ฅ^8 ) (1/(1 + ๐ฅ) " + " 2๐ฅ/(1 + ๐ฅ^2 ) " + " (4๐ฅ^3)/(1 + ๐ฅ^4 ) " +" (8๐ฅ^7)/(1 + ๐ฅ^8 )) Hence, ๐โฒ(๐) = (๐+๐)(๐+๐^๐ )(๐+๐^๐ )(๐+๐^๐ ) (๐/(๐ + ๐) " + " ๐๐/(๐ + ๐^๐ ) " + " (๐๐^๐)/(๐ + ๐^๐ ) " + " (๐๐^๐)/(๐ + ๐^๐ )) We need to find ๐โฒ(1) Putting ๐ฅ=1 ๐โฒ(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)