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

Go Ad-free

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) ))

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo