Ex 5.5, 17 - Chapter 5 Class 12 Continuity and Differentiability

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Ex 5.5, 17 Differentiate ( 𝑥﷮2﷯ – 5𝑥 + 8) ( 𝑥﷮3﷯ + 7𝑥 + 9) (𝑖) By using product rule Let y = ( 𝑥﷮2﷯ – 5𝑥 + 8) ( 𝑥﷮3﷯ + 7𝑥 + 9) By using product rule 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑥﷮2﷯ − 5𝑥 + 8﷯ 𝑥﷮3﷯ + 7𝑥 + 9﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 𝑥﷮2﷯ – 5𝑥 + 8﷯ ﷮𝑑𝑥﷯ . 𝑥﷮3﷯+7𝑥+9﷯ + 𝑑 𝑥﷮3﷯+ 7𝑥 + 9﷯﷮𝑑𝑥﷯ . 𝑥﷮2﷯ – 5𝑥 + 8﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2𝑥−5+0﷯ 𝑥﷮3﷯+7𝑥+9﷯ + 3 𝑥﷮2﷯+7+0﷯ 𝑥﷮2﷯ – 5𝑥 + 8﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2𝑥−5﷯ 𝑥﷮3﷯+7𝑥+9﷯ + 3 𝑥﷮2﷯+7﷯ 𝑥﷮2﷯ – 5𝑥 + 8﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2𝑥 𝑥﷮3﷯+7𝑥+9﷯−5 𝑥﷮3﷯+7𝑥+9﷯ + 3 𝑥﷮2﷯ 𝑥﷮2﷯ – 5𝑥 + 8﷯ + 7 𝑥﷮2﷯ – 5𝑥 + 8﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2 𝑥﷮4﷯+14 𝑥﷮2﷯+18𝑥−5 𝑥﷮3﷯ − 35𝑥−45+3 𝑥﷮4﷯ −15𝑥 + 24 𝑥﷮2﷯+7 𝑥﷮2﷯−35𝑥+5 𝒅𝒚﷮𝒅𝒙﷯ = 𝟓 𝒙﷮𝟒﷯−𝟐𝟎 𝒙﷮𝟑﷯+ 𝟒𝟓 𝒙﷮𝟐﷯−𝟓𝟐𝒙+𝟏𝟏 Ex 5.5, 17 Differentiate ( 𝑥﷮2﷯– 5 𝑥 + 8) ( 𝑥﷮3﷯ + 7 𝑥 + 9) (ii) by expanding the product to obtain a single polynomial. By Expanding the product to obtain a single polynomial . 𝑦= 𝑥﷮2﷯– 5 𝑥 + 8﷯ 𝑥﷮3﷯ + 7 𝑥 + 9﷯ 𝑦= 𝑥﷮2﷯ 𝑥﷮3﷯ + 7 𝑥 + 9﷯ – 5𝑥 𝑥﷮3﷯ + 7 𝑥 + 9﷯ 8 𝑥﷮3﷯ + 7 𝑥 + 9﷯ 𝑦= 𝑥﷮5﷯+7 𝑥﷮3﷯+9 𝑥﷮2﷯−5 𝑥﷮4﷯−35 𝑥﷮2﷯−45𝑥+8 𝑥﷮3﷯+56𝑥+72 𝑦= 𝑥﷮5﷯−5 𝑥﷮4﷯+15 𝑥﷮3﷯−26 𝑥﷮2﷯+11𝑥+72 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑( 𝑥﷮5﷯ − 5 𝑥﷮4﷯ + 15 𝑥﷮3﷯− 26 𝑥﷮2﷯ + 11𝑥 + 72 ) ﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑( 𝑥﷮5﷯)﷮𝑑𝑥﷯ − 𝑑(5 𝑥﷮4﷯)﷮𝑑𝑥﷯ + 𝑑(15 𝑥﷮3﷯) ﷮𝑑𝑥﷯ − 𝑑(26 𝑥﷮2﷯) ﷮𝑑𝑥﷯ + 𝑑(11𝑥) ﷮𝑑𝑥﷯ + 𝑑(72) ﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 5 𝑥﷮4﷯−20 𝑥﷮3﷯+45 𝑥﷮2﷯−52𝑥+11 + 0 𝒅𝒚﷮𝒅𝒙﷯ = 𝟓 𝒙﷮𝟒﷯−𝟐𝟎 𝒙﷮𝟑﷯+𝟒𝟓 𝒙﷮𝟐﷯−𝟓𝟐𝒙+𝟏𝟏 Ex 5.5, 17 Differentiate ( 𝑥﷮2﷯– 5 𝑥 + 8) ( 𝑥﷮3﷯ + 7 𝑥 + 9) (iii) by logarithmic differentiation. By logarithmic differentiation 𝑦= 𝑥﷮2﷯– 5 𝑥 + 8﷯ 𝑥﷮3﷯ + 7 𝑥 + 9﷯ Taking log both sides log 𝑦 = log 𝑥﷮2﷯– 5 𝑥 + 8﷯ 𝑥﷮3﷯ + 7 𝑥 + 9﷯﷯ log 𝑦=log 𝑥﷮2﷯– 5 𝑥 + 8﷯+ log ﷮ 𝑥﷮3﷯ + 7 𝑥 + 9﷯﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑 log﷮𝑦﷯﷯ ﷮𝑑𝑥﷯ = 𝑑 log 𝑥﷮2﷯ – 5𝑥 + 8﷯ + log ﷮ 𝑥﷮3﷯ + 7𝑥 + 9﷯﷯﷯﷮𝑑𝑥﷯ 𝑑 log﷮𝑦﷯﷯ ﷮𝑑𝑥﷯ . 𝑑𝑦﷮𝑑𝑦﷯ = 𝑑 log 𝑥﷮2﷯ – 5𝑥 + 8﷯﷯﷮𝑑𝑥﷯ + 𝑑 log ﷮ 𝑥﷮3﷯ + 7𝑥 + 9﷯﷯﷯﷮𝑑𝑥﷯ 𝑑 log﷮𝑦﷯﷯ ﷮𝑑𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮ 𝑥﷮2﷯ – 5𝑥 + 8﷯﷯ . 𝑑 𝑥﷮2﷯ – 5𝑥 + 8﷯﷮𝑑𝑥﷯ + 1﷮ 𝑥﷮3﷯ + 7𝑥 + 9﷯﷯ . 𝑑 𝑥﷮3﷯ + 7𝑥 + 9﷯﷮𝑑𝑥﷯ 1 ﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮ 𝑥﷮2﷯ – 5𝑥 + 8﷯ . (2x – 5 + 0) + 1﷮ 𝑥﷮3﷯ + 7𝑥 + 9﷯ .(3x2 + 7 + 0) 1 ﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ = 2𝑥 − 5﷯﷮ 𝑥﷮2﷯ – 5𝑥 + 8﷯ + 3 𝑥﷮2﷯ + 7﷯﷮ 𝑥﷮3﷯ + 7𝑥 + 9﷯ 1 ﷮𝑦﷯ . 𝑑𝑦﷮𝑑𝑥﷯ = 2𝑥 − 5﷯ 𝑥﷮3﷯ + 7𝑥 + 9﷯ + 3 𝑥﷮2﷯ + 7﷯ 𝑥﷮2﷯ – 5𝑥 + 8﷯﷮ 𝑥﷮2﷯ – 5𝑥 + 8﷯ 𝑥﷮3﷯ + 7𝑥 + 9﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑦 2𝑥 − 5﷯ 𝑥﷮3﷯ + 7𝑥 + 9﷯ + 3 𝑥﷮2﷯ + 7﷯ 𝑥﷮2﷯ – 5𝑥 + 8﷯﷮ 𝑥﷮2﷯ – 5𝑥 + 8﷯ 𝑥﷮3﷯ + 7𝑥 + 9﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑥﷮2﷯ – 5𝑥 + 8﷯ 𝑥﷮3﷯ + 7𝑥 + 9﷯ 2𝑥 − 5﷯ 𝑥﷮3﷯ + 7𝑥 + 9﷯ + 3 𝑥﷮2﷯ + 7﷯ 𝑥﷮2﷯ – 5𝑥 + 8﷯﷮ 𝑥﷮2﷯ – 5𝑥 + 8﷯ 𝑥﷮3﷯ + 7𝑥 + 9﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2𝑥−5﷯ 𝑥﷮3﷯+7𝑥+9﷯ + 3 𝑥﷮2﷯+7﷯ 𝑥﷮2﷯ – 5𝑥 + 8﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2𝑥 𝑥﷮3﷯+7𝑥+9﷯−5 𝑥﷮3﷯+7𝑥+9﷯ + 3 𝑥﷮2﷯ 𝑥﷮2﷯ – 5𝑥 + 8﷯ + 7 𝑥﷮2﷯ – 5𝑥 + 8﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2 𝑥﷮4﷯+14 𝑥﷮2﷯+18𝑥−5 𝑥﷮3﷯ − 35𝑥−45+3 𝑥﷮4﷯ −15𝑥 + 24 𝑥﷮2﷯+7 𝑥﷮2﷯−35𝑥+5 𝒅𝒚﷮𝒅𝒙﷯ = 𝟓 𝒙﷮𝟒﷯−𝟐𝟎 𝒙﷮𝟑﷯+ 𝟒𝟓 𝒙﷮𝟐﷯−𝟓𝟐𝒙+𝟏𝟏 Hence, the answer is same in all three cases .