Logarithmic Differentiation - Type 2

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Misc 10 Differentiate w.r.t. x the function, π₯π₯ + π₯π + π^π₯+ ππ, for some fixed π >0 and π₯> 0Let π¦= π₯π₯ + π₯π + π^π₯+ ππ And let u=π₯π₯ , π£=π₯π , π€=π^π₯ Now, π=π+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(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π€/ππ₯ + 0 ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ + ππ€/ππ₯ Calculating ππ/ππ π’ =π₯^π₯ Taking log on both sides logβ‘π’=logβ‘γπ₯^π₯ γ logβ‘π’=π₯ .logβ‘π₯ (π^π ππ  ππππ π‘πππ‘) Differentiating both sides π€.π.π‘.π₯. π(logβ‘π’ )/ππ₯ = π(π₯ .γ logγβ‘π₯ )/ππ₯ π(logβ‘π’ )/ππ₯ . ππ’/ππ’ = π(π₯ .γ logγβ‘π₯ )/ππ₯ π(logβ‘π’ )/ππ’ . ππ’/ππ₯ = ππ₯/ππ₯ . logβ‘π₯ + (π(γ logγβ‘π₯))/ππ₯. π₯ 1/π’ . ππ’/ππ₯ = logβ‘π₯ + 1/π₯ . π₯ 1/π’ . ππ’/ππ₯ = logβ‘π₯ + 1 ππ’/ππ₯ = u (1+ log π₯)β‘ ππ/ππ = π^π (π+ πππ π)β‘ Calculating ππ/π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 βπ) Calculating ππ/π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₯ .logβ‘π Therefore, 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^(π βπ) + π^π .πππβ‘π