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

Last updated at April 16, 2024 by Teachoo

Ex 5.5

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Ex 5.5, 5 You are here

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Last updated at April 16, 2024 by Teachoo

Ex 5.5, 5 Differentiate the functions in, (𝑥 + 3)^2 . (𝑥 + 4)^3 . (𝑥 + 5)^4 Let 𝑦= (𝑥 + 3)^2 . (𝑥 + 4)^3 . (𝑥 + 5)^4 Taking log both sides log𝑦 = log ((𝑥 + 3)^2 . (𝑥 + 4)^3 . (𝑥 + 5)^4 ) log𝑦 = log (𝑥 + 3)^2 + log (𝑥 + 4)^3 + log (𝑥 + 5)^4 log𝑦 = 2 log (𝑥 + 3) + 3 log (𝑥 + 4) + 4 log (𝑥 + 5) Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑(log𝑦 )/𝑑𝑥 = (𝑑 (2 log (𝑥 + 3)" + " 3 log (𝑥 + 4)" + " 4 log (𝑥 + 5)))/𝑑𝑥 𝑑(log𝑦 )/𝑑𝑥 (𝑑𝑦/𝑑𝑦) = 𝑑(2 log (𝑥 + 3))/𝑑𝑥 + (𝑑 (3 log (𝑥 + 4)))/𝑑𝑥 + 𝑑(4 log (𝑥 + 5))/𝑑𝑥 𝑑(log𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑𝑥) = 2 𝑑(log (𝑥 + 3))/𝑑𝑥 + 3 (𝑑 (log (𝑥 + 4)))/𝑑𝑥 + 4 𝑑(log (𝑥 + 5))/𝑑𝑥 1/𝑦 × 𝑑𝑦/𝑑𝑥 = 2. 1/((𝑥 + 3) ) . 𝑑(𝑥 + 3)/𝑑𝑥 + 3. 1/((𝑥 + 4) ) . 𝑑(𝑥 + 4)/𝑑𝑥 + 4. 1/((𝑥 + 5) ) . 𝑑(𝑥 + 5)/𝑑𝑥 1/𝑦 × 𝑑𝑦/𝑑𝑥 = 2/(𝑥 + 3) (𝑑𝑥/𝑑𝑥+𝑑(3)/𝑑𝑥) + 3/(𝑥 + 4) (𝑑𝑥/𝑑𝑥+𝑑(4)/𝑑𝑥) + 4/(𝑥 +5) (𝑑𝑥/𝑑𝑥+𝑑(5)/𝑑𝑥) 1/𝑦 × 𝑑𝑦/𝑑𝑥 = 2/(𝑥 + 3) (1+0) + 3/(𝑥 + 4) (1+0) + 4/(𝑥 + 5) (1+0) 1/𝑦 × 𝑑𝑦/𝑑𝑥 = 2/(𝑥 + 3) + 3/(𝑥 + 4) + 4/(𝑥 + 5) 𝑑𝑦/𝑑𝑥 = 𝑦 (2/(𝑥 + 3) " + " 3/(𝑥 + 4) " + " 4/(𝑥 + 5)) Putting value of 𝑦 𝑑𝑦/𝑑𝑥 = (𝑥 + 3)^2 . (𝑥 + 4)^3 . (𝑥 + 5)^(4 ) (2/((𝑥 + 3) ) "+ " 3/((𝑥 + 4) ) " + " 4/((𝑥 + 5) )) 𝑑𝑦/𝑑𝑥 = (𝑥 + 3)^2 (𝑥 + 4)^3 (𝑥 + 5)^(4 ) ((2(𝑥 + 4) (𝑥 + 5) + 3(𝑥 + 3) (𝑥 + 5) + 4(𝑥 + 3) (𝑥 + 4))/((𝑥 + 3) (𝑥 + 4) (𝑥 + 5) )) 𝑑𝑦/𝑑𝑥 = ((𝑥 + 3)^2 (𝑥 + 4)^3 〖 (𝑥 + 5)〗^(4 ))/((𝑥 + 3) (𝑥 + 4) (𝑥 + 5) ) (2(𝑥^2+4𝑥+5𝑥+20)+3(𝑥^2+3𝑥+5𝑥+15)+ 4(𝑥^2+3𝑥+4𝑥+12)) 𝑑𝑦/𝑑𝑥 =(𝑥 + 3) (𝑥 + 4)^2 〖 (𝑥 + 5)〗^(3 ) (2(𝑥^2+9𝑥+20)+3(𝑥^2+8𝑥+15)+4(𝑥^2+7𝑥+12)) 𝑑𝑦/𝑑𝑥 =(𝑥 + 3) (𝑥 + 4)^2 〖 (𝑥 + 5)〗^(3 ) (2𝑥^2+18𝑥+40+3𝑥^2+24𝑥+45+4𝑥^2+28𝑥+48) 𝑑𝑦/𝑑𝑥 =(𝑥 + 3) (𝑥 + 4)^2 〖 (𝑥 + 5)〗^(3 ) (2𝑥^2+3𝑥^2+4𝑥^2 18𝑥+24𝑥+28𝑥+40+45+48) 𝒅𝒚/𝒅𝒙 =(𝒙 + 𝟑) (𝒙 + 𝟒)^𝟐 〖 (𝒙 + 𝟓)〗^(𝟑 ) (𝟗𝒙^𝟐+𝟕𝟎𝒙+𝟏𝟑𝟑)