Get live Maths 1-on-1 Classs - Class 6 to 12
Example 30 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at March 22, 2023 by Teachoo
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Transcript
Example 30 Differentiate β(((π₯β3) (π₯2+4))/( 3π₯2+ 4π₯ + 5)) π€.π.π‘. π₯.Let y =β(((π₯ β 3) (π₯^2 + 4))/( 3π₯2+ 4π₯ + 5)) Taking log on both sides πππβ‘π = πππ β(((π β π) (ππ + π))/( πππ+ ππ + π)) logβ‘π¦ = logβ‘ (((π₯ β 3) (π₯2 + 4))/( 3π₯2+ 4π₯ + 5))^(1/2) πππβ‘π = π/π log ((π₯ β 3) (π₯2 + 4))/((3π₯2+ 4π₯ + 5) ) (ππ πππ logβ‘γπ^π γ=π logβ‘π) πππβ‘π = π/π log ((π₯ β 3) (π₯2 + 4))/((3π₯2+ 4π₯ + 5) ) πππβ‘π¦ = 1/2 (logβ‘γ (π₯β3)γ+γlog γβ‘(π₯2 + 4)βlogβ‘γ (3π₯2 + 4π₯ + 5)γ ) Differentiating π€.π.π‘.π₯ (π (logβ‘π¦))/ππ₯ = 1/2 ((π(logβ‘(π₯β3)+γlog γβ‘γ(π₯^2+4)βlogβ‘γ (3π₯^2+4π₯+5)γ γ ) )/ππ₯) (π (logβ‘π¦))/ππ₯ = 1/2 ((π(logβ‘(π₯β3)))/ππ₯ " + " π(logβ‘(π₯^2+4) )/ππ₯ " β " π(logβ‘(3π₯^2+4π₯+5) )/ππ₯) 1/π¦ (ππ¦/ππ₯) = 1/2 [1/(π₯ β 3) " . " π(π₯ β3)/ππ₯ " + " 1/(π₯^2 + 4) " . " (π (π₯^2 + 4))/ππ₯ " β " 1/((3π₯^2+ 4π₯+ 5)) " . " (π (3π₯^2+ 4π₯+ 5))/(ππ₯ )] ππ πππ πππβ‘ππ=πππβ‘π+πππ β‘π &πππβ‘γπ/πγ=πππβ‘γπβπππβ‘π γ 1/π¦ (ππ¦/ππ₯) = 1/2 [1/((π₯ β 3) ) " " (1β0) "+ " 1/(π₯^2+ 4) " " (2π₯ + 0)" β " 1/(3π₯^2+ 4π₯ + 5) " " (6π₯ +4+0)" " ] 1/π¦ (ππ¦/ππ₯) = 1/2 (1/((π₯ β 3) )+2π₯/(π₯^2+ 4)β(6π₯ + 4)/(3π₯^2 + 4π₯ + 5)) ππ¦/ππ₯ = π¦/2 (1/((π₯ β 3) )+2π₯/(π₯^2+ 4)β(6π₯ + 4)/(3π₯^2 + 4π₯ + 5)) π π/π π = π/π β(((π β π)(π^π+ π))/(ππ^π+ ππ + π)) (π/((π β π) )+ππ/(π^π+ π)β(ππ + π)/(ππ^π + ππ + π))
