Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month
Logarithmic Differentiation - Type 2
Last updated at March 11, 2021 by Teachoo
Ex 5.5, 4 Differentiate the functions in, π₯^π₯ β 2^sinβ‘π₯ Let π¦=π₯^π₯ β 2^sinβ‘π₯ Let π’=π₯^π₯ , π£=2^sinβ‘π₯ π¦= π’βπ£ Differentiating both sides π€.π.π‘.π₯. ππ¦/ππ₯ = π(π’ β π£)/ππ₯ ππ¦/ππ₯ = ππ’/ππ₯ β ππ£/ππ₯ Calculating π π/π π π’=π₯^π₯ Taking log both sides log π’=γlogβ‘γ (π₯γγ^π₯) log π’= π₯ . logβ‘π₯ Differentiating both sides π€.π.π‘.π₯. π(logβ‘π’ )/ππ₯ = π(π₯ . logβ‘π₯ )/ππ₯ π(logβ‘π’ )/ππ₯ (ππ’/ππ’) = π(π₯ . logβ‘π₯ )/ππ₯ π(logβ‘π’ )/ππ’ (ππ’/ππ₯) = π(π₯ . logβ‘π₯ )/ππ₯ v(As πππβ‘(π^π) = π πππβ‘π) 1/π’ . ππ’/ππ₯ = π(π₯ . logβ‘π₯ )/ππ₯ 1/π’. ππ’/ππ₯ = (π π₯ )/ππ₯ .logβ‘π₯ + (π (logβ‘π₯ ))/ππ₯ . π₯ 1/π’ . ππ’/ππ₯ = 1 . logβ‘π₯ + 1/π₯ . π₯ 1/π’ . ππ’/ππ₯ = logβ‘γπ₯+1γ ππ’/ππ₯ = π’ (logβ‘γπ₯+1γ ) ππ’/ππ₯ = π₯^π₯ (logβ‘γπ₯+1γ ) Using product value in γπ₯ .γβ‘πππβ‘π₯ As (π’π£)β = π’βπ£ + π£βπ’ Calculating π π/π π π£=2^sinβ‘π₯ Taking log both sides log π£=logβ‘γ (2^sinβ‘π₯ ) " " γ log π£= sin π₯ . logβ‘2 Differentiating both sides π€.π.π‘.π₯. π(logβ‘π£ )/ππ₯ = π(sin π₯" . " logβ‘2 )/ππ₯ π(logβ‘π£ )/ππ₯ (ππ£/ππ£) = logβ‘2 π(sin π₯)/ππ₯ π(logβ‘π£ )/ππ£ (ππ£/ππ₯) = logβ‘2 . cosβ‘π₯ (As πππβ‘(π^π) = π πππβ‘π) 1/π£ (ππ£/ππ₯) = logβ‘2 . cosβ‘π₯ ππ£/ππ₯ = π¦ (logβ‘2 . cosβ‘π₯) ππ£/ππ₯ = 2^sinβ‘π₯ (logβ‘2 . cosβ‘π₯) Now ππ¦/ππ₯ = ππ’/ππ₯ β ππ£/ππ₯ Putting values of ππ’/ππ₯ & ππ£/ππ₯ ππ¦/ππ₯ = π₯^π₯ (logβ‘γπ₯+1γ ) β2^sinβ‘π₯ (logβ‘2" . " cosβ‘π₯) π π/π π = π^π (π+πππβ‘π ) βπ^πππβ‘π (ππ¨π¬β‘π.πππβ‘π)