Logarithmic Differentiation - Type 2

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month

### Transcript

Ex 5.5, 9 Differentiate the functions in, π₯^sinβ‘π₯ + γ(sinβ‘π₯)γ^cosβ‘π₯ Let y = π₯^sinβ‘π₯ + γ(sinβ‘π₯)γ^cosβ‘γπ₯ γ Let π’ =π₯^sinβ‘π₯ & π£ =γ(sinβ‘π₯)γ^cosβ‘π₯ 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₯^sinβ‘π₯ Taking log both sides logβ‘π’=logβ‘γπ₯^sinβ‘π₯ γ logβ‘π’= sinβ‘π₯. logβ‘γ π₯γ Differentiating both sides π€.π.π‘.π₯. (π(logβ‘π’))/ππ₯ = π(sinβ‘π₯. logβ‘γ π₯γ )/ππ₯ (π(logβ‘π’))/ππ₯ . ππ’/ππ’ = π(sinβ‘π₯. logβ‘γ π₯γ )/ππ₯ 1/π’ . ππ’/ππ₯ = π(sinβ‘π₯ . logβ‘γ π₯γ )/ππ₯ (As πππβ‘(π^π) = π . πππβ‘π) Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ 1/π’ . ππ’/ππ₯ = π(sinβ‘π₯ )/ππ₯ . logβ‘π₯ + π(logβ‘π₯ )/ππ₯ . sin π₯ 1/π’ . ππ’/ππ₯ = cosβ‘π₯ . logβ‘π₯ + 1/π₯ . sin π₯ 1/π’ . ππ’/ππ₯ = cosβ‘π₯ . logβ‘π₯ + (sin π₯" " )/π₯ ππ’/ππ₯ = u(cosβ‘γπ₯ .logβ‘γπ₯+ sinβ‘π₯/π₯γ γ ) ππ’/ππ₯ = π₯^sinβ‘π₯ (cosβ‘γπ₯ .logβ‘γπ₯+ sinβ‘π₯/π₯γ γ ). ππ’/ππ₯ = π₯^sinβ‘π₯ (γsinβ‘π₯/π₯ +cosγβ‘γπ₯ .logβ‘π₯ γ ) Calculating ππ/ππ π£= (sinβ‘π₯ )^cosβ‘π₯ Taking log both sides logβ‘π£ = logβ‘γ γ. sinγβ‘π₯γ^(cos π₯) logβ‘π£=cosβ‘π₯. log sinβ‘π₯ Differentiating both sides π€.π.π‘.π₯. (π(logβ‘π£))/ππ₯ = π(cosβ‘π₯. log sinβ‘π₯ )/ππ₯ (π(logβ‘π£))/ππ₯ . ππ£/ππ£ = π(cosβ‘π₯. log sinβ‘π₯ )/ππ₯ (π(logβ‘π£))/ππ£ . ππ£/ππ₯ = π(cosβ‘π₯. log sinβ‘π₯ )/ππ₯ (As πππβ‘(π^π) = π . πππβ‘π) 1/π£ . ππ£/ππ₯ = π(cosβ‘π₯. log sinβ‘π₯ )/ππ₯ 1/π£ . ππ£/ππ₯ = π(cosβ‘π₯ )/ππ₯ . logβ‘sinβ‘π₯ + π(logβ‘sinβ‘π₯ )/ππ₯ . cosβ‘π₯ 1/π£ . ππ£/ππ₯ = βsinβ‘π₯ . logβ‘sinβ‘π₯ + (1/sinβ‘π₯ . π(sinβ‘π₯ )/ππ₯) . cosβ‘π₯ 1/π£ . ππ£/ππ₯ = βsinβ‘π₯ . logβ‘sinβ‘π₯ + (1/sinβ‘π₯ .cosβ‘π₯ ) . cosβ‘π₯ 1/π£ . ππ£/ππ₯ = βsinβ‘π₯ . logβ‘sinβ‘π₯ + (cotβ‘π₯ ) . cosβ‘π₯ ππ£/ππ₯ = π£(βsinβ‘π₯ " " . logβ‘sinβ‘π₯" +" cotβ‘π₯ ". " cosβ‘π₯) ππ£/ππ₯ = (sinβ‘π₯ )^cosβ‘π₯ (coπ β‘π₯ ". " coπ‘β‘π₯βsinβ‘π₯ " " . logβ‘sinβ‘π₯) using product Rule As (π’π£)β = π’βπ£ + π£βπ’ 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^π¬π’π§β‘π (π¬π’π§β‘π/π+ππ¨π¬β‘γπ .π₯π¨π β‘π γ ) + (π¬π’π§β‘π )^ππ¨π¬β‘π (ππ¨π¬β‘π .ππ¨π­β‘πβπ¬π’π§β‘π π₯π¨π β‘π¬π’π§β‘π )