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

Last updated at Jan. 3, 2020 by Teachoo

Last updated at Jan. 3, 2020 by Teachoo

Transcript

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 (As ๐๐๐โก(๐๐)= ๐๐๐โก๐ + ๐๐๐โก๐) 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) = 16 ((1 + 2 + 4 + 8)/2) = 16 ร 15/2 = 8 "ร " 15 = 120

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