Ex 5.4, 6 - Ex 5.4

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Teachoo ยท Class 12 Explore Class 12

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Ex 5.4, 6 Differentiate ๐‘ค.๐‘Ÿ.๐‘ก. ๐‘ฅ in , ๐‘’^๐‘ฅ+ ๐‘’^(๐‘ฅ^2 ) + ๐‘’^(๐‘ฅ^3 )+ ๐‘’^(๐‘ฅ^4 ) + ๐‘’^(๐‘ฅ^5 )Let ๐‘ฆ = ๐‘’^๐‘ฅ+ ๐‘’^(๐‘ฅ^2 ) +... + ๐‘’^(๐‘ฅ^5 ) y = ๐‘’^๐‘ฅ+ ๐‘’^(๐‘ฅ^2 ) + ๐‘’^(๐‘ฅ^3 )+ ๐‘’^(๐‘ฅ^4 ) + ๐‘’^(๐‘ฅ^5 ) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘(๐‘ฆ)/๐‘‘๐‘ฅ = ๐‘‘(๐‘’^๐‘ฅ " + " ๐‘’^(๐‘ฅ^2 ) " + " ๐‘’^(๐‘ฅ^3 ) " + " ๐‘’^(๐‘ฅ^4 ) " + " ๐‘’^(๐‘ฅ^5 ) )/๐‘‘๐‘ฅ = (๐‘‘(๐‘’^๐‘ฅ ) )/๐‘‘๐‘ฅ + ๐‘‘(๐‘’^(๐‘ฅ^2 ) )/๐‘‘๐‘ฅ + ๐‘‘(๐‘’^(๐‘ฅ^3 ) )/๐‘‘๐‘ฅ + ๐‘‘(๐‘’^(๐‘ฅ^4 ) )/๐‘‘๐‘ฅ + ๐‘‘(๐‘’^(๐‘ฅ^5 ) )/๐‘‘๐‘ฅ = ๐‘’^๐‘ฅ + ๐‘’^(๐‘ฅ^2 ). ๐‘‘(๐‘ฅ^2 )/๐‘‘๐‘ฅ + ๐‘’^(๐‘ฅ^3 ). ๐‘‘(๐‘ฅ^3 )/๐‘‘๐‘ฅ + ๐‘’^(๐‘ฅ^4 ). ๐‘‘(๐‘ฅ^4 )/๐‘‘๐‘ฅ + ๐‘’^(๐‘ฅ^5 ). ๐‘‘(๐‘ฅ^5 )/๐‘‘๐‘ฅ = ๐’†^๐’™ + 2x๐’†^(๐’™^๐Ÿ ) + ๐Ÿ‘๐’™^๐Ÿ ๐’†^(๐’™^๐Ÿ‘ ) + ๐Ÿ’๐’™^๐Ÿ‘ ๐’†^(๐’™^๐Ÿ’ ) + ใ€–๐Ÿ“๐’™ใ€—^(๐Ÿ’ ) ๐’†^(๐’™^๐Ÿ“ )

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