Ex 5.4, 6 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 5.4, 6 Differentiate ๐ค.๐.๐ก. ๐ฅ in , ๐^๐ฅ+ ๐^(๐ฅ^2 ) + ๐^(๐ฅ^3 )+ ๐^(๐ฅ^4 ) + ๐^(๐ฅ^5 )Let ๐ฆ = ๐^๐ฅ+ ๐^(๐ฅ^2 ) +... + ๐^(๐ฅ^5 ) y = ๐^๐ฅ+ ๐^(๐ฅ^2 ) + ๐^(๐ฅ^3 )+ ๐^(๐ฅ^4 ) + ๐^(๐ฅ^5 ) Differentiating both sides ๐ค.๐.๐ก.๐ฅ ๐(๐ฆ)/๐๐ฅ = ๐(๐^๐ฅ " + " ๐^(๐ฅ^2 ) " + " ๐^(๐ฅ^3 ) " + " ๐^(๐ฅ^4 ) " + " ๐^(๐ฅ^5 ) )/๐๐ฅ = (๐(๐^๐ฅ ) )/๐๐ฅ + ๐(๐^(๐ฅ^2 ) )/๐๐ฅ + ๐(๐^(๐ฅ^3 ) )/๐๐ฅ + ๐(๐^(๐ฅ^4 ) )/๐๐ฅ + ๐(๐^(๐ฅ^5 ) )/๐๐ฅ = ๐^๐ฅ + ๐^(๐ฅ^2 ). ๐(๐ฅ^2 )/๐๐ฅ + ๐^(๐ฅ^3 ). ๐(๐ฅ^3 )/๐๐ฅ + ๐^(๐ฅ^4 ). ๐(๐ฅ^4 )/๐๐ฅ + ๐^(๐ฅ^5 ). ๐(๐ฅ^5 )/๐๐ฅ = ๐^๐ + 2x๐^(๐^๐ ) + ๐๐^๐ ๐^(๐^๐ ) + ๐๐^๐ ๐^(๐^๐ ) + ใ๐๐ใ^(๐ ) ๐^(๐^๐ )
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