Logarithmic Differentiation - Type 2

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Ex 5.5, 6 Differentiate the functions in, (π₯+1/π₯)^π₯+ π₯^((1 + 1/π₯) ) Let π¦= (π₯+1/π₯)^π₯+ π₯^((1 + 1/π₯) ) Let π’ = (π₯+1/π₯)^π₯ , π£ = π₯^((1 + 1/π₯) ) 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₯+1/π₯)^π₯ Taking log both sides logβ‘π’ = log (π₯+1/π₯)^π₯ logβ‘π’ = π₯ log (π₯+1/π₯) Differentiating both sides π€.π.π‘.π₯. π(logβ‘π’ )/ππ₯ = (π (π₯ log" " (π₯ + 1/π₯)))/ππ₯ π(logβ‘π’ )/ππ₯ (ππ’/ππ’) = (π (π₯ log" " (π₯ + 1/π₯)))/ππ₯ (As πππβ‘(π^π )=π . πππβ‘π) π(logβ‘π’ )/ππ’ (ππ’/ππ₯)" = " (π (π₯ log" " (π₯ + 1/π₯)))/ππ₯ 1/π’ (ππ’/ππ₯)" = " (π (π₯ log" " (π₯ + 1/π₯)))/ππ₯ 1/π’ (ππ’/ππ₯)" = " π(π₯)/ππ₯ . log (π₯ + 1/π₯) + π(log" " (π₯ + 1/π₯))/ππ₯ . π₯ 1/π’ (ππ’/ππ₯)" =" 1. log (π₯ + 1/π₯) + ((1/(π₯ + 1/π₯)).π/ππ₯ (π₯ + 1/π₯)) . π₯ 1/π’ (ππ’/ππ₯)" =" log (π₯ + 1/π₯) + (1/(π₯ + 1/π₯) . (π(π₯)/ππ₯+(π (1/π₯))/ππ₯)) . π₯ Using product rule in π₯ πππ" " (π₯ + 1/π₯) As (uv)β = uβ v + vβ u 1/π’ (ππ’/ππ₯)" =" log (π₯+1/π₯) + (1/(π₯ + 1/π₯) . (1+(β1)/π₯^2 " " )) . π₯ 1/π’ (ππ’/ππ₯)" =" log (π₯+1/π₯) + (π₯/(π₯^2 + 1) . (1β1/π₯^2 " " )) . π₯ 1/π’ (ππ’/ππ₯)" =" log (π₯+1/π₯) + (π₯/(π₯^2 + 1) ((π₯^2 β 1)/π₯^2 ).π₯) 1/π’ (ππ’/ππ₯)" =" log (π₯+1/π₯) + (π₯/(π₯^2 + 1) ((π₯^2 β 1)/π₯^2 ).π₯) 1/π’ (ππ’/ππ₯)" =" log (π₯+1/π₯) + (π₯^2/π₯^2 ((π₯^2 β 1)/(π₯^2 + 1))) 1/π’ (ππ’/ππ₯)" =" log (π₯+1/π₯) + ((π₯^2 β 1)/(π₯^2+ 1)) ππ’/ππ₯ "= " π’ (γlog γβ‘(π₯+1/π₯)+((π₯^2 β 1)/(π₯^2+ 1))) ππ’/ππ₯ "=" (π₯+1/π₯)^π₯ (γlog γβ‘(π₯+1/π₯)+((π₯^2 β 1)/(π₯^2+ 1))) ππ/ππ "=" (π+π/π)^π ((π^π β π)/(π^π+ π)β‘γ+ γπππ γβ‘(π+π/π) γ ) Calculating ππ/ππ π£ = π₯^(1 + 1/π₯)" " Taking log both sides log π£ = log π₯^(1 + 1/π₯)" " log π£ = (1 + 1/π₯)log π₯^" " Differentiating both sides π€.π.π‘.π₯. π(logβ‘π£ )/ππ₯ = (π ((1 + 1/π₯)" . " log π₯))/ππ₯ π(logβ‘π£ )/ππ₯ (ππ£/ππ£) = (π ((1 + 1/π₯)" . " log π₯))/ππ₯ π(logβ‘π£ )/ππ£ (ππ£/ππ₯) = (π ((1 + 1/π₯)" . " log π₯))/ππ₯ 1/π£ (ππ£/ππ₯) = (π ((1 + 1/π₯)" . " log π₯))/ππ₯ Using product rule in (π₯+ 1/π₯)" . " πππ π₯ 1/π£ (ππ£/ππ₯) = π(1 + 1/π₯)/ππ₯ . logβ‘π₯ + π(logβ‘π₯ )/ππ₯ . (1 + 1/π₯) 1/π£ (ππ£/ππ₯) = (π(1)/ππ₯+π(1/π₯)/ππ₯) . logβ‘π₯ + 1/π₯ (1 + 1/π₯) 1/π£ (ππ£/ππ₯) = (0+((β1)/π₯^2 )) . logβ‘π₯ + 1/π₯ (1 + 1/π₯) 1/π£ (ππ£/ππ₯) = (β1)/π₯^2 . logβ‘π₯ + 1/π₯ (1 + 1/π₯) 1/π£ (ππ£/ππ₯) = (βlogβ‘π₯)/π₯^2 + 1/π₯ + 1/π₯^2 1/π£ (ππ£/ππ₯) = (βlogβ‘π₯)/π₯^2 + 1/π₯ + 1/π₯^2 1/π£ (ππ£/ππ₯) = ((βlogβ‘π₯ + π₯ + 1)/π₯^2 ) ππ£/ππ₯ = π£ ((βlogβ‘π₯ + π₯ + 1)/π₯^2 ) ππ£/ππ₯ = π₯^((1 + 1/π₯) ) ((π₯ + 1 β logβ‘π₯ )/π₯^2 ) Now ππ£/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Putting values of ππ’/ππ₯ & ππ£/ππ₯ ππ/ππ = (π+π/π)^π ((π^π β π)/(π^π+ π)+π₯π¨π β‘(π+ π/π) ) + π^((π + π/π) ) ((π + π β πππβ‘π )/π^π )