Logarithmic Differentiation - Type 2

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Ex 5.5, 10 Differentiate the functions in, π₯^(π₯ πππ β‘π₯ ) + (π₯2+ 1)/(π₯2β 1)Let y = π₯^(π₯ πππ β‘π₯ ) + (π₯2+ 1)/(π₯2β 1) Let π’ =π₯^(π₯ πππ β‘π₯ ) & π£ =(π₯2+ 1)/(π₯2β 1) β΄ π¦ = π’+π£ Differentiating both sides π€.π.π‘.π₯. ππ¦/ππ₯ = (π (π’ + π£))/ππ₯ ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Calculating ππ/ππ π’ =π₯^(π₯ πππ β‘π₯ ) Taking log both sides logβ‘π’=logβ‘γπ₯^(π₯ πππ β‘π₯ ) γ logβ‘π’=π₯ πππ β‘π₯. logβ‘γ π₯γ Differentiating both sides π€.π.π‘.π₯. (π(logβ‘π’))/ππ₯ = π(π₯ πππ β‘π₯ logβ‘π₯ )/ππ₯ (π(logβ‘π’))/ππ₯ . ππ’/ππ’ = π(π₯ πππ β‘π₯ logβ‘π₯ )/ππ₯ (π(logβ‘π’))/ππ’ . ππ’/ππ₯ = π(π₯ πππ β‘π₯ logβ‘π₯ )/ππ₯ (As πππβ‘(π^π) = π . πππβ‘π) 1/π’ . ππ’/ππ₯ = π(π₯ πππ β‘π₯ logβ‘π₯ )/ππ₯ 1/π’ . ππ’/ππ₯ = π(π₯ cosβ‘π₯ )/ππ₯ logβ‘π₯ + π₯ πππ β‘π₯ (π(logβ‘π₯))/ππ₯ 1/π’ . ππ’/ππ₯ = π(π₯ cosβ‘π₯ )/ππ₯ logβ‘π₯ + π₯ πππ β‘π₯ Γ1/π₯ 1/π’ . ππ’/ππ₯ = [π(π₯)/ππ₯ cosβ‘π₯+π₯ π(cosβ‘π₯ )/ππ₯]logβ‘π₯ + πππ β‘π₯ 1/π’ . ππ’/ππ₯ = [cosβ‘π₯βπ₯ π ππ π₯]logβ‘π₯ + πππ β‘π₯ 1/π’ . ππ’/ππ₯ = cosβ‘π₯ . logβ‘π₯ β π₯ sin π₯ logβ‘π₯+cosβ‘π₯ Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ Where u = x cos x, v = log x 1/π’ . ππ’/ππ₯ = cosβ‘π₯ . logβ‘π₯+cosβ‘π₯β π₯ sin π₯ logβ‘π₯ 1/π’ ππ’/ππ₯ = cosβ‘π₯ (logβ‘π₯+1)βπ₯ sinβ‘γπ₯ logβ‘π₯ γ ππ’/ππ₯ = u (cosβ‘π₯ (logβ‘π₯+1)βπ₯ sinβ‘γπ₯ logβ‘π₯ γ ) ππ’/ππ₯ = π₯^(π₯ πππ β‘π₯ ) (cosβ‘γ (logβ‘γπ₯+1γ )βπ₯ sinβ‘γπ₯ logβ‘π₯ γ γ ) Calculating ππ/ππ π£= (π₯2 + 1)/(π₯2 β 1) Differentiating both sides π€.π.π‘.π₯. π(π£)/ππ₯ = π((π₯2 + 1)/(π₯2 β 1))/ππ₯ π(π£)/ππ₯ . ππ£/ππ£ = π((π₯2 + 1)/(π₯2 β 1))/ππ₯ π(π£)/ππ£ . ππ£/ππ₯ = π((π₯2 + 1)/(π₯2 β 1))/ππ₯ 1. ππ£/ππ₯ = π((π₯2 + 1)/(π₯2 β 1))/ππ₯ ππ£/ππ₯ = (π(π₯2+ 1)/ππ₯ . (π₯^2 β 1) β π(π₯^2 β 1)/ππ₯ . (π₯2+ 1))/(π₯^2 β 1)^2 ππ£/ππ₯ = ((2π₯ + 0) (π₯^2 β1) β (2π₯ β 0) (π₯^2 + 1))/(π₯^2 β1)^2 Using quotient rule (π’/π£)β² = (π’^β² π£ β π£^β² π’)/π£^2 ππ£/ππ₯ = (2π₯ (π₯^2 β1) β 2π₯ (π₯^2 + 1))/(π₯^2 β1)^2 ππ£/ππ₯ = (2π₯ (π₯^2 β1 βπ₯^2 β1))/(π₯^2 β1)^2 ππ£/ππ₯ = (2π₯ (β 2))/(π₯^2 β1)^2 ππ£/ππ₯ = (β4π₯)/(π₯^2 β1)^2 Now ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Putting value of ππ’/ππ₯ & ππ£/ππ₯ ππ/ππ = π^γπ ππ¨π¬γβ‘π (ππ¨π¬β‘γπ (π+π₯π¨π β‘π ) βπ π¬π’π§β‘γπ π₯π¨π β‘π γ γ ) β ππ/(π^π βπ)^π