Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

Last updated at March 11, 2021 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

Transcript

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₯ ) logβ‘π¦ = log β‘(cosβ‘π₯) + log β‘(2π₯) + log β‘(cosβ‘3π₯) Differentiating both sides π€.π.π‘.π₯. π(logβ‘π¦ )/ππ₯ = π(log β‘(cosβ‘π₯)" + " log β‘(2π₯) "+ " log β‘(cosβ‘3π₯))/ππ₯ π(logβ‘π¦ )/ππ₯ (ππ¦/ππ¦) = (π(log β‘(cosβ‘π₯)) )/ππ₯ + (π(log β‘(2π₯)) )/ππ₯ + (π(log β‘(cosβ‘3π₯)) )/ππ₯ π(logβ‘π¦ )/ππ¦ (ππ¦/ππ₯) = 1/cosβ‘π₯ . (π (cosβ‘π₯ ))/ππ₯ + 1/cosβ‘2π₯ . (π(cosβ‘2π₯))/ππ₯ + 1/cosβ‘3π₯ . π(cosβ‘3π₯ )/ππ₯ 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₯ 1/π¦ . ππ¦/ππ₯ = (βsinβ‘π₯)/cosβ‘π₯ β sinβ‘2π₯/cosβ‘π₯ . 2 β sinβ‘3π₯/cosβ‘3π₯ . 3 1/π¦ . ππ¦/ππ₯ = βtanβ‘π₯βtanβ‘2π₯. 2 βtanβ‘3π₯. 3 1/π¦ . ππ¦/ππ₯ = β (tanβ‘π₯+2 tanβ‘2π₯+3 tanβ‘3π₯ ) 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 )

Ex 5.5

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.