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Ex 5.5, 17 - Differentiate using product rule, by expanding product

Ex 5.5, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.5, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.5, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.5, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 5

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Transcript

Ex 5.5, 17 Differentiate (𝑥^2 – 5𝑥 + 8) (𝑥^3 + 7𝑥 + 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" )" – 5" 𝑥(𝑥^3 " + 7" 𝑥" + 9" )" + 8 " (𝑥^3 " + 7" 𝑥" + 9" ) 𝑦=𝑥^5+7𝑥^3+9𝑥^2−5𝑥^4−35𝑥^2−45𝑥+8𝑥^3+56𝑥+72 𝑦=𝑥^5−5𝑥^4+15𝑥^3−26𝑥^2+11𝑥+72 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦/𝑑𝑥 = (𝑑(𝑥^5 − 5𝑥^4 + 15𝑥^3− 26𝑥^2 + 11𝑥 + 72" " )" " )/𝑑𝑥 𝑑𝑦/𝑑𝑥 = (𝑑(𝑥^5))/𝑑𝑥 − (𝑑(5𝑥^4))/𝑑𝑥 + (𝑑(15𝑥^3)" " )/𝑑𝑥 − (𝑑(26𝑥^2)" " )/𝑑𝑥 + (𝑑(11𝑥)" " )/𝑑𝑥 + (𝑑(72)" " )/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 5𝑥^4−20𝑥^3+45𝑥^2−52𝑥+11 + 0 𝒅𝒚/𝒅𝒙 = 𝟓𝒙^𝟒−𝟐𝟎𝒙^𝟑+𝟒𝟓𝒙^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) + ((3𝑥^2 + 7))/(𝑥^3 " + " 7𝑥" +" 9) (1 )/𝑦 . 𝑑𝑦/𝑑𝑥 = ((2𝑥 − 5) (𝑥^3 " + " 7𝑥" +" 9) + (3𝑥^2 + 7) (𝑥^2 " – " 5𝑥" + " 8))/((𝑥^2 " – " 5𝑥" + " 8) (𝑥^3 " + " 7𝑥" +" 9) ) 𝑑𝑦/𝑑𝑥 = 𝑦(((2𝑥 − 5) (𝑥^3 " + " 7𝑥" +" 9) + (3𝑥^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) + (3𝑥^2 + 7) (𝑥^2 " – " 5𝑥" + " 8))/((𝑥^2 " – " 5𝑥" + " 8) (𝑥^3 " + " 7𝑥" +" 9) )) (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) + ((3𝑥^2 + 7))/(𝑥^3 " + " 7𝑥" +" 9) (1 )/𝑦 . 𝑑𝑦/𝑑𝑥 = ((2𝑥 − 5) (𝑥^3 " + " 7𝑥" +" 9) + (3𝑥^2 + 7) (𝑥^2 " – " 5𝑥" + " 8))/((𝑥^2 " – " 5𝑥" + " 8) (𝑥^3 " + " 7𝑥" +" 9) ) 𝑑𝑦/𝑑𝑥 = 𝑦(((2𝑥 − 5) (𝑥^3 " + " 7𝑥" +" 9) + (3𝑥^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) + (3𝑥^2 + 7) (𝑥^2 " – " 5𝑥" + " 8))/((𝑥^2 " – " 5𝑥" + " 8) (𝑥^3 " + " 7𝑥" +" 9) )) (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) + ((3𝑥^2 + 7))/(𝑥^3 " + " 7𝑥" +" 9) (1 )/𝑦 . 𝑑𝑦/𝑑𝑥 = ((2𝑥 − 5) (𝑥^3 " + " 7𝑥" +" 9) + (3𝑥^2 + 7) (𝑥^2 " – " 5𝑥" + " 8))/((𝑥^2 " – " 5𝑥" + " 8) (𝑥^3 " + " 7𝑥" +" 9) ) 𝑑𝑦/𝑑𝑥 = 𝑦(((2𝑥 − 5) (𝑥^3 " + " 7𝑥" +" 9) + (3𝑥^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) + (3𝑥^2 + 7) (𝑥^2 " – " 5𝑥" + " 8))/((𝑥^2 " – " 5𝑥" + " 8) (𝑥^3 " + " 7𝑥" +" 9) )) (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) + ((3𝑥^2 + 7))/(𝑥^3 " + " 7𝑥" +" 9) (1 )/𝑦 . 𝑑𝑦/𝑑𝑥 = ((2𝑥 − 5) (𝑥^3 " + " 7𝑥" +" 9) + (3𝑥^2 + 7) (𝑥^2 " – " 5𝑥" + " 8))/((𝑥^2 " – " 5𝑥" + " 8) (𝑥^3 " + " 7𝑥" +" 9) ) 𝑑𝑦/𝑑𝑥 = 𝑦(((2𝑥 − 5) (𝑥^3 " + " 7𝑥" +" 9) + (3𝑥^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) + (3𝑥^2 + 7) (𝑥^2 " – " 5𝑥" + " 8))/((𝑥^2 " – " 5𝑥" + " 8) (𝑥^3 " + " 7𝑥" +" 9) ))

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.