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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.5, 17 Differentiate (๐‘ฅ^2 โ€“ 5๐‘ฅ + 8) (๐‘ฅ^3 + 7๐‘ฅ + 9) (๐‘–) By using product rule Let y = (๐‘ฅ^2 โ€“ 5๐‘ฅ + 8) (๐‘ฅ^3 + 7๐‘ฅ + 9) By using product rule ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= ๐‘‘((๐‘ฅ^2 โˆ’ 5๐‘ฅ + 8)(๐‘ฅ^3 + 7๐‘ฅ + 9))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(๐‘ฅ^2 " โ€“ " 5๐‘ฅ" +" 8)" " /๐‘‘๐‘ฅ . (๐‘ฅ^3+7๐‘ฅ+9) + ๐‘‘" " (๐‘ฅ^3+ 7๐‘ฅ + 9)/๐‘‘๐‘ฅ . (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" +" 8) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (2๐‘ฅโˆ’5+0) (๐‘ฅ^3+7๐‘ฅ+9) + (3๐‘ฅ^2+7+0) (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" +" 8) Using product Rule As (๐‘ข๐‘ฃ)โ€™ = ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (2๐‘ฅโˆ’5) (๐‘ฅ^3+7๐‘ฅ+9) + (3๐‘ฅ^2+7) (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" +" 8) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2๐‘ฅ (๐‘ฅ^3+7๐‘ฅ+9)โˆ’5 (๐‘ฅ^3+7๐‘ฅ+9) + 3๐‘ฅ^2 (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" +" 8) + 7 (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" +" 8) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2๐‘ฅ^4+14๐‘ฅ^2+18๐‘ฅโˆ’5๐‘ฅ^3 โˆ’ 35๐‘ฅโˆ’ 45+3๐‘ฅ^4 โˆ’15๐‘ฅ+ 24๐‘ฅ^2+7๐‘ฅ^2โˆ’35๐‘ฅ+5 ๐’…๐’š/๐’…๐’™ = ๐Ÿ“๐’™^๐Ÿ’โˆ’๐Ÿ๐ŸŽ๐’™^๐Ÿ‘+ ๐Ÿ’๐Ÿ“๐’™^๐Ÿโˆ’๐Ÿ“๐Ÿ๐’™+๐Ÿ๐Ÿ Ex 5.5, 17 Differentiate (๐‘ฅ^2 โ€“ 5๐‘ฅ + 8) (๐‘ฅ^3 + 7๐‘ฅ + 9) (ii) by expanding the product to obtain a single polynomial. By Expanding the product to obtain a single polynomial . ๐‘ฆ=(๐‘ฅ^2 " โ€“ 5" ๐‘ฅ" + 8" ) (๐‘ฅ^3 " + 7" ๐‘ฅ" + 9" ) ๐‘ฆ=๐‘ฅ^2 (๐‘ฅ^3 " + 7" ๐‘ฅ" + 9" )" โ€“ 5" ๐‘ฅ(๐‘ฅ^3 " + 7" ๐‘ฅ" + 9" )" + 8 " (๐‘ฅ^3 " + 7" ๐‘ฅ" + 9" ) ๐‘ฆ=๐‘ฅ^5+7๐‘ฅ^3+9๐‘ฅ^2โˆ’5๐‘ฅ^4โˆ’35๐‘ฅ^2โˆ’45๐‘ฅ+8๐‘ฅ^3+56๐‘ฅ+72 ๐‘ฆ=๐‘ฅ^5โˆ’5๐‘ฅ^4+15๐‘ฅ^3โˆ’26๐‘ฅ^2+11๐‘ฅ+72 Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘(๐‘ฅ^5 โˆ’ 5๐‘ฅ^4 + 15๐‘ฅ^3โˆ’ 26๐‘ฅ^2 + 11๐‘ฅ + 72" " )" " )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘(๐‘ฅ^5))/๐‘‘๐‘ฅ โˆ’ (๐‘‘(5๐‘ฅ^4))/๐‘‘๐‘ฅ + (๐‘‘(15๐‘ฅ^3)" " )/๐‘‘๐‘ฅ โˆ’ (๐‘‘(26๐‘ฅ^2)" " )/๐‘‘๐‘ฅ + (๐‘‘(11๐‘ฅ)" " )/๐‘‘๐‘ฅ + (๐‘‘(72)" " )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 5๐‘ฅ^4โˆ’20๐‘ฅ^3+45๐‘ฅ^2โˆ’52๐‘ฅ+11 + 0 ๐’…๐’š/๐’…๐’™ = ๐Ÿ“๐’™^๐Ÿ’โˆ’๐Ÿ๐ŸŽ๐’™^๐Ÿ‘+๐Ÿ’๐Ÿ“๐’™^๐Ÿโˆ’๐Ÿ“๐Ÿ๐’™+๐Ÿ๐Ÿ Ex 5.5, 17 Differentiate (๐‘ฅ^2โ€“ 5 ๐‘ฅ + 8) (๐‘ฅ^3 + 7 ๐‘ฅ + 9) (iii) by logarithmic differentiation. By logarithmic differentiation ๐‘ฆ= (๐‘ฅ^2 "โ€“ 5 " ๐‘ฅ" + 8" ) (๐‘ฅ^3 " + 7 " ๐‘ฅ" + 9" ) Taking log both sides log ๐‘ฆ = log ((๐‘ฅ^2 " โ€“ 5" ๐‘ฅ" + 8" ) (๐‘ฅ^3 " + 7" ๐‘ฅ" + 9" )) log ๐‘ฆ=log (๐‘ฅ^2 " โ€“ 5" ๐‘ฅ" + 8" )+ใ€–log ใ€—โก(๐‘ฅ^3 " + 7" ๐‘ฅ" + 9" ) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. (๐‘‘(logโก๐‘ฆ ) )/๐‘‘๐‘ฅ = ๐‘‘(log (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8) + ใ€–log ใ€—โก(๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) )/๐‘‘๐‘ฅ By logarithmic differentiation ๐‘ฆ= (๐‘ฅ^2 "โ€“ 5 " ๐‘ฅ" + 8" ) (๐‘ฅ^3 " + 7 " ๐‘ฅ" + 9" ) Taking log both sides log ๐‘ฆ = log ((๐‘ฅ^2 " โ€“ 5" ๐‘ฅ" + 8" ) (๐‘ฅ^3 " + 7" ๐‘ฅ" + 9" )) log ๐‘ฆ=log (๐‘ฅ^2 " โ€“ 5" ๐‘ฅ" + 8" )+ใ€–log ใ€—โก(๐‘ฅ^3 " + 7" ๐‘ฅ" + 9" ) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. (๐‘‘(logโก๐‘ฆ ) )/๐‘‘๐‘ฅ = ๐‘‘(log (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8) + ใ€–log ใ€—โก(๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) )/๐‘‘๐‘ฅ (As log (ab) = log a + log b) (๐‘‘(logโก๐‘ฆ ) )/๐‘‘๐‘ฅ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ = ๐‘‘(log (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8))/๐‘‘๐‘ฅ + ๐‘‘(ใ€–log ใ€—โก(๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) )/๐‘‘๐‘ฅ (๐‘‘(logโก๐‘ฆ ) )/๐‘‘๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/((๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8) ) . ๐‘‘(๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8)/๐‘‘๐‘ฅ + 1/((๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) ) . ๐‘‘(๐‘ฅ^3 " + " 7๐‘ฅ" +" 9)/๐‘‘๐‘ฅ (1 )/๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/(๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8) . (2x โ€“ 5 + 0) + 1/(๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) .(3x2 + 7 + 0) (1 )/๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((2๐‘ฅ โˆ’ 5))/(๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8) + ((3๐‘ฅ^2 + 7))/(๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) (1 )/๐‘ฆ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((2๐‘ฅ โˆ’ 5) (๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) + (3๐‘ฅ^2 + 7) (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8))/((๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8) (๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฆ(((2๐‘ฅ โˆ’ 5) (๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) + (3๐‘ฅ^2 + 7) (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8))/((๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8) (๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) )) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ =(๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8) (๐‘ฅ^3 " + " 7๐‘ฅ" +" 9)(((2๐‘ฅ โˆ’ 5) (๐‘ฅ^3 " + " 7๐‘ฅ" + " 9) + (3๐‘ฅ^2 + 7) (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8))/((๐‘ฅ^2 " โ€“ " 5๐‘ฅ" + " 8) (๐‘ฅ^3 " + " 7๐‘ฅ" +" 9) )) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (2๐‘ฅโˆ’5) (๐‘ฅ^3+7๐‘ฅ+9) + (3๐‘ฅ^2+7) (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" +" 8) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2๐‘ฅ (๐‘ฅ^3+7๐‘ฅ+9)โˆ’5 (๐‘ฅ^3+7๐‘ฅ+9) + 3๐‘ฅ^2 (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" +" 8) + 7 (๐‘ฅ^2 " โ€“ " 5๐‘ฅ" +" 8) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2๐‘ฅ^4+14๐‘ฅ^2+18๐‘ฅโˆ’5๐‘ฅ^3 โˆ’ 35๐‘ฅโˆ’45+3๐‘ฅ^4 โˆ’15๐‘ฅ+ 24๐‘ฅ^2+7๐‘ฅ^2โˆ’35๐‘ฅ+5 ๐’…๐’š/๐’…๐’™ = ๐Ÿ“๐’™^๐Ÿ’โˆ’๐Ÿ๐ŸŽ๐’™^๐Ÿ‘+ ๐Ÿ’๐Ÿ“๐’™^๐Ÿโˆ’๐Ÿ“๐Ÿ๐’™+๐Ÿ๐Ÿ Hence, the answer is same in all three cases .

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.