Logarithmic Differentiation - Type 2

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Misc 11 Differentiate w.r.t. x the function, π₯^(π₯^2β 3)+(π₯β3)π₯^2, for π₯ > 3Let π¦=π₯^(π₯^2β 3)+(π₯β3)^(π₯^2 ) And let π’=π₯^(π₯^2β 3) , π£ =(π₯β3)^(π₯^2 ) Now, 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₯^2β 3) Taking log on both sides log π’=logβ‘γπ₯^(π₯^2β 3) γ log π’=γ(π₯γ^2β 3). logβ‘π₯ Differentiating π€.π.π‘.π₯. π(logβ‘π’ )/ππ₯ = π(γ(π₯γ^2β 3) logβ‘π₯ )/ππ₯ π(logβ‘π’ )/ππ₯ . ππ’/ππ’ = π(γ(π₯γ^2 β 3) logβ‘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(γ(π₯γ^2β 3) logβ‘π₯ )/ππ₯ " " 1/π’ . ππ’/ππ₯ = π(γ(π₯γ^2β 3) logβ‘π₯ )/ππ₯ 1/π’ . ππ’/ππ₯ = (πγ(π₯γ^2β 3) )/ππ₯ . γ logγβ‘π₯ + π(logβ‘π₯ )/ππ₯ . γ(π₯γ^2β 3) 1/π’ . ππ’/ππ₯ = (2π₯ β0) γ logγβ‘π₯ + 1/π₯ Γ γ(π₯γ^2β 3) 1/π’ . ππ’/ππ₯ = 2π₯ . logβ‘π₯ + (π₯^2β 3)/π₯ ππ’/ππ₯ = u (2π₯ "." logβ‘π₯ "+ " (π₯^2β 3)/π₯) Using Product rule As (π’π£)β = π’βπ£ + π£βπ’ ππ/ππ = π^(π^πβ π) (ππ "." πππβ‘π "+ " (π^πβ π)/π) Calculating ππ/ππ π£ = (π₯β3)π₯^2 Taking log on both sides log π£=logβ‘γ(π₯β3)^(π₯^2 ) γ log π£=γπ₯^2 . logγβ‘γ (π₯β3)γ Differentiating π€.π.π‘.π₯. π(logβ‘π£ )/ππ₯ = (π(γπ₯^2. logγβ‘γ (π₯ β 3)γ ) )/ππ₯ (As logβ‘(π^π) = π logβ‘π) π(logβ‘π£ )/ππ₯ . ππ£/ππ£ = (π(γπ₯^2. logγβ‘γ (π₯β3)γ ) )/ππ₯ π(logβ‘π£ )/ππ£ . ππ£/ππ₯ = (π(γπ₯^2. logγβ‘γ (π₯β3)γ ) )/ππ₯ 1/π£ . ππ£/ππ₯ = (π(γπ₯^2. logγβ‘γ (π₯β3)γ ) )/ππ₯ 1/π£ . ππ£/ππ₯ = π(π₯^2 )/ππ₯ . log (π₯β3) + π(log" " (π₯ β 3))/ππ₯ . π₯^2 1/π£ . ππ£/ππ₯ = 2π₯ . log (π₯β3) + 1/((π₯ β 3) ). (π(π₯ β 3)" " )/ππ₯ . π₯^2 1/π£ . ππ£/ππ₯ = 2π₯ . log (π₯β3) + 1/((π₯ β 3) ) . π₯^2 1/π£ . ππ£/ππ₯ = 2π₯. log (π₯β3) + π₯^2/(π₯ β3)Using product rule (π’π£)β = π’βπ£ + π£βπ’ ππ£/ππ₯ = π£ (2π₯". " log" " (π₯β3)" + " π₯^2/(π₯ β3)) ππ/ππ = (πβπ)π^π (ππ". " π₯π¨π " " (πβπ)" + " π^π/(π βπ)) Now, ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ = π^(π^πβ π) ((π^πβ π)/π+ππ π₯π¨π β‘π ) + (πβπ)π^π (π^π/(π βπ)+ππ .π₯π¨π β‘(π βπ) )