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Ex 5.5

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Last updated at Dec. 7, 2021 by Teachoo

Ex 5.5, 1 Differentiate the functions in, cosβ‘π₯ . cosβ‘2π₯ . cosβ‘3π₯ Let y = cosβ‘π₯ . cosβ‘2π₯ . cosβ‘3π₯ Taking log both sides logβ‘π¦ = log (cosβ‘π₯.cosβ‘2π₯.cosβ‘3π₯ ) πππβ‘π = πππ β‘(πππβ‘π) + πππ β‘(πππ ππ) + πππ β‘(πππβ‘ππ) Differentiating both sides π€.π.π‘.π₯. π(logβ‘π¦ )/ππ₯ = π(log β‘(cosβ‘π₯)" + " log β‘(cosβ‘2π₯) "+ " log β‘(cosβ‘3π₯))/ππ₯ π(logβ‘π¦ )/ππ₯ (ππ¦/ππ¦) = (π(log β‘(cosβ‘π₯)) )/ππ₯ + (π(log β‘(cosβ‘2π₯)) )/ππ₯ + (π(log β‘(cosβ‘3π₯)) )/ππ₯ π (πππβ‘π )/π π (π π/π π) = π/ππ¨π¬β‘π . (π (ππ¨π¬β‘π ))/π π + π/ππ¨π¬β‘ππ . (π (ππ¨π¬β‘ππ))/π π + π/ππ¨π¬β‘ππ . π (ππ¨π¬β‘ππ )/π π 1/π¦ . ππ¦/ππ₯ = 1/cosβ‘π₯ .(β sinβ‘π₯) + 1/cosβ‘2π₯ .(β sinβ‘2π₯).π(2π₯)/ππ₯ + 1/cosβ‘π₯ .(β sinβ‘3π₯).π(3π₯)/ππ₯ π/π . π π/π π = (βπ¬π’π§β‘π)/ππ¨π¬β‘π β π¬π’π§β‘ππ/ππ¨π¬β‘π . π β π¬π’π§β‘ππ/ππ¨π¬β‘ππ . π 1/π¦ . ππ¦/ππ₯ = βtanβ‘π₯βtanβ‘2π₯. 2 βtanβ‘3π₯. 3 π/π . π π/π π = β (πππβ‘π+π πππβ‘ππ+π πππβ‘ππ ) ππ¦/ππ₯ = βπ¦ (tanβ‘π₯+2 tanβ‘2π₯+3 tanβ‘3π₯ ) π π/π π = β πππβ‘π . πππβ‘ππ . πππβ‘ππ (πππβ‘π+π πππβ‘ππ+π πππβ‘ππ )