# Ex 5.5, 3 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by Teachoo

Last updated at March 11, 2021 by Teachoo

Transcript

Ex 5.5, 3 Differentiate the functions in, (logβ‘π₯ )^cosβ‘π₯ Let π¦=(logβ‘π₯ )^cosβ‘π₯ Taking log both sides logβ‘π¦ = logβ‘γγ (logβ‘π₯ )γ^cosβ‘π₯ γ logβ‘π¦ = cosβ‘γπ₯ .γlog γβ‘(logβ‘π₯ ) γ Differentiating both sides π€.π.π‘.π₯. π(logβ‘π¦)/ππ₯ = π(cosβ‘γπ₯ .γ log γβ‘(logβ‘π₯ ) γ )/ππ₯ π(logβ‘π¦ )/ππ₯ (ππ¦/ππ¦) = π(cosβ‘γπ₯ .γ log γβ‘(logβ‘π₯ ) γ )/ππ₯ π(logβ‘π¦ )/ππ₯ (ππ¦/ππ₯) = π(cosβ‘γπ₯ .γ log γβ‘(logβ‘π₯ ) γ )/ππ₯ (As πππβ‘(π^π) = π πππβ‘π) 1/π¦ . ππ¦/ππ₯ = π(cosβ‘γπ₯ .γ log γβ‘(logβ‘π₯ ) γ )/ππ₯ 1/π¦ ππ¦/ππ₯ = π(cosβ‘π₯ )/ππ₯ . γ log γβ‘(logβ‘π₯ ) + π(γ log γβ‘(logβ‘π₯ ) )/ππ₯ . cosβ‘π₯ 1/π¦ ππ¦/ππ₯ = γβsinγβ‘π₯ . γlog γβ‘(logβ‘π₯ ) + 1/logβ‘π₯ . π(logβ‘π₯ )/ππ₯ . cosβ‘π₯ 1/π¦ ππ¦/ππ₯ = γβsinγβ‘π₯ . γlog γβ‘(logβ‘π₯ ) + 1/logβ‘π₯ Γ 1/π₯ . cosβ‘π₯ 1/π¦ ππ¦/ππ₯ = γβsinγβ‘π₯ . γlog γβ‘(logβ‘π₯ ) + cosβ‘π₯/(π₯ logβ‘π₯ ) ππ¦/ππ₯ = π¦ (γβsinγβ‘π₯ " . " γlog γβ‘(logβ‘π₯ )" + " cosβ‘π₯/(π₯ logβ‘π₯ )) Using product rule in πππ β‘γπ₯ .γ πππ γβ‘(πππβ‘π₯ ) γ (π’π£)β = π’βπ£ + π£βπ’ Putting values of π¦ ππ¦/ππ₯ = (πππβ‘π₯ )^πππ β‘π₯ (γβπ ππγβ‘π₯ " . " γπππ γβ‘(πππβ‘π₯ )" + " πππ β‘π₯/(π₯ πππβ‘π₯ )) π π/π π = (πππβ‘π )^πππβ‘π (πππβ‘π/(π πππβ‘π ) γ β πππγβ‘π " . " γπ₯π¨π γβ‘(πππβ‘π ) )

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.