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

Last updated at March 11, 2021 by Teachoo

Last updated at March 11, 2021 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 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)

Ex 5.5

Ex 5.5, 1
Important

Ex 5.5, 2

Ex 5.5, 3 Important

Ex 5.5, 4

Ex 5.5, 5

Ex 5.5,6 Important

Ex 5.5, 7 Important

Ex 5.5, 8

Ex 5.5, 9 Important

Ex 5.5, 10 Important

Ex 5.5, 11 Important

Ex 5.5, 12

Ex 5.5, 13

Ex 5.5, 14 Important

Ex 5.5, 15

Ex 5.5, 16 Important You are here

Ex 5.5, 17 Important

Ex 5.5, 18

Chapter 5 Class 12 Continuity and Differentiability

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.