Ex 5.5
Last updated at August 25, 2026 by Teachoo
Transcript
Ex 5.5, 17 Differentiate (π₯^2 β 5π₯ + 8) (π₯^3 + 7π₯ + 9) (ii) by expanding the product to obtain a single polynomial.By Expanding the product to obtain a single polynomial . π¦=(π₯^2 " β 5" π₯" + 8" ) (π₯^3 " + 7" π₯" + 9" ) π¦=π₯^2 (π₯^3 " + 7" π₯" + 9" )" β 5" π₯(π₯^3 " + 7" π₯" + 9" )" + 8 " (π₯^3 " + 7" π₯" + 9" ) π¦=π₯^5+7π₯^3+9π₯^2β5π₯^4β35π₯^2β45π₯+8π₯^3+56π₯+72 π¦=π₯^5β5π₯^4+15π₯^3β26π₯^2+11π₯+72 Differentiating both sides π€.π.π‘.π₯. ππ¦/ππ₯ = (π(π₯^5 β 5π₯^4 + 15π₯^3β 26π₯^2 + 11π₯ + 72" " )" " )/ππ₯ ππ¦/ππ₯ = (π(π₯^5))/ππ₯ β (π(5π₯^4))/ππ₯ + (π(15π₯^3)" " )/ππ₯ β (π(26π₯^2)" " )/ππ₯ + (π(11π₯)" " )/ππ₯ + (π(72)" " )/ππ₯ ππ¦/ππ₯ = 5π₯^4β20π₯^3+45π₯^2β52π₯+11 + 0 π π/π π = ππ^πβπππ^π+πππ^πβπππ+ππ Ex 5.5, 17 Differentiate (π₯^2β 5 π₯ + 8) (π₯^3 + 7 π₯ + 9) (iii) by logarithmic differentiation.By logarithmic differentiation π¦= (π₯^2 "β 5 " π₯" + 8" ) (π₯^3 " + 7 " π₯" + 9" ) Taking log both sides log π¦ = log ((π₯^2 " β 5" π₯" + 8" ) (π₯^3 " + 7" π₯" + 9" )) log π¦=log (π₯^2 " β 5" π₯" + 8" )+γlog γβ‘(π₯^3 " + 7" π₯" + 9" ) Differentiating both sides π€.π.π‘.π₯. (π(logβ‘π¦ ) )/ππ₯ = π(log (π₯^2 " β " 5π₯" + " 8) + γlog γβ‘(π₯^3 " + " 7π₯" +" 9) )/ππ₯ (π(logβ‘π¦ ) )/ππ₯ . ππ¦/ππ¦ = π(log (π₯^2 " β " 5π₯" + " 8))/ππ₯ + π(γlog γβ‘(π₯^3 " + " 7π₯" +" 9) )/ππ₯ (π(logβ‘π¦ ) )/ππ¦ . ππ¦/ππ₯ = 1/((π₯^2 " β " 5π₯" + " 8) ) . π(π₯^2 " β " 5π₯" + " 8)/ππ₯ + 1/((π₯^3 " + " 7π₯" +" 9) ) . π(π₯^3 " + " 7π₯" +" 9)/ππ₯ (1 )/π¦ . ππ¦/ππ₯ = 1/(π₯^2 " β " 5π₯" + " 8) . (2x β 5 + 0) + 1/(π₯^3 " + " 7π₯" +" 9) .(3x2 + 7 + 0) (1 )/π¦ . ππ¦/ππ₯ = ((2π₯ β 5))/(π₯^2 " β " 5π₯" + " 8) + ((3π₯^2 + 7))/(π₯^3 " + " 7π₯" +" 9) (1 )/π¦ . ππ¦/ππ₯ = ((2π₯ β 5) (π₯^3 " + " 7π₯" +" 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) ) ππ¦/ππ₯ = π¦(((2π₯ β 5) (π₯^3 " + " 7π₯" +" 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) )) ππ¦/ππ₯ =(π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9)(((2π₯ β 5) (π₯^3 " + " 7π₯" + " 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) )) (π(logβ‘π¦ ) )/ππ₯ . ππ¦/ππ¦ = π(log (π₯^2 " β " 5π₯" + " 8))/ππ₯ + π(γlog γβ‘(π₯^3 " + " 7π₯" +" 9) )/ππ₯ (π(logβ‘π¦ ) )/ππ¦ . ππ¦/ππ₯ = 1/((π₯^2 " β " 5π₯" + " 8) ) . π(π₯^2 " β " 5π₯" + " 8)/ππ₯ + 1/((π₯^3 " + " 7π₯" +" 9) ) . π(π₯^3 " + " 7π₯" +" 9)/ππ₯ (1 )/π¦ . ππ¦/ππ₯ = 1/(π₯^2 " β " 5π₯" + " 8) . (2x β 5 + 0) + 1/(π₯^3 " + " 7π₯" +" 9) .(3x2 + 7 + 0) (1 )/π¦ . ππ¦/ππ₯ = ((2π₯ β 5))/(π₯^2 " β " 5π₯" + " 8) + ((3π₯^2 + 7))/(π₯^3 " + " 7π₯" +" 9) (1 )/π¦ . ππ¦/ππ₯ = ((2π₯ β 5) (π₯^3 " + " 7π₯" +" 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) ) ππ¦/ππ₯ = π¦(((2π₯ β 5) (π₯^3 " + " 7π₯" +" 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) )) ππ¦/ππ₯ =(π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9)(((2π₯ β 5) (π₯^3 " + " 7π₯" + " 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) )) (π(logβ‘π¦ ) )/ππ₯ . ππ¦/ππ¦ = π(log (π₯^2 " β " 5π₯" + " 8))/ππ₯ + π(γlog γβ‘(π₯^3 " + " 7π₯" +" 9) )/ππ₯ (π(logβ‘π¦ ) )/ππ¦ . ππ¦/ππ₯ = 1/((π₯^2 " β " 5π₯" + " 8) ) . π(π₯^2 " β " 5π₯" + " 8)/ππ₯ + 1/((π₯^3 " + " 7π₯" +" 9) ) . π(π₯^3 " + " 7π₯" +" 9)/ππ₯ (1 )/π¦ . ππ¦/ππ₯ = 1/(π₯^2 " β " 5π₯" + " 8) . (2x β 5 + 0) + 1/(π₯^3 " + " 7π₯" +" 9) .(3x2 + 7 + 0) (1 )/π¦ . ππ¦/ππ₯ = ((2π₯ β 5))/(π₯^2 " β " 5π₯" + " 8) + ((3π₯^2 + 7))/(π₯^3 " + " 7π₯" +" 9) (1 )/π¦ . ππ¦/ππ₯ = ((2π₯ β 5) (π₯^3 " + " 7π₯" +" 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) ) ππ¦/ππ₯ = π¦(((2π₯ β 5) (π₯^3 " + " 7π₯" +" 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) )) ππ¦/ππ₯ =(π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9)(((2π₯ β 5) (π₯^3 " + " 7π₯" + " 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) )) (π(logβ‘π¦ ) )/ππ₯ . ππ¦/ππ¦ = π(log (π₯^2 " β " 5π₯" + " 8))/ππ₯ + π(γlog γβ‘(π₯^3 " + " 7π₯" +" 9) )/ππ₯ (π(logβ‘π¦ ) )/ππ¦ . ππ¦/ππ₯ = 1/((π₯^2 " β " 5π₯" + " 8) ) . π(π₯^2 " β " 5π₯" + " 8)/ππ₯ + 1/((π₯^3 " + " 7π₯" +" 9) ) . π(π₯^3 " + " 7π₯" +" 9)/ππ₯ (1 )/π¦ . ππ¦/ππ₯ = 1/(π₯^2 " β " 5π₯" + " 8) . (2x β 5 + 0) + 1/(π₯^3 " + " 7π₯" +" 9) .(3x2 + 7 + 0) (1 )/π¦ . ππ¦/ππ₯ = ((2π₯ β 5))/(π₯^2 " β " 5π₯" + " 8) + ((3π₯^2 + 7))/(π₯^3 " + " 7π₯" +" 9) (1 )/π¦ . ππ¦/ππ₯ = ((2π₯ β 5) (π₯^3 " + " 7π₯" +" 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) ) ππ¦/ππ₯ = π¦(((2π₯ β 5) (π₯^3 " + " 7π₯" +" 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) )) ππ¦/ππ₯ =(π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9)(((2π₯ β 5) (π₯^3 " + " 7π₯" + " 9) + (3π₯^2 + 7) (π₯^2 " β " 5π₯" + " 8))/((π₯^2 " β " 5π₯" + " 8) (π₯^3 " + " 7π₯" +" 9) ))