Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity

Last updated at Jan. 3, 2020 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity

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

Ex 5.5, 1
Important
You are here

Ex 5.5, 2

Ex 5.5, 3 Important

Ex 5.5, 4

Ex 5.5, 5

Ex 5.5,6 Important

Ex 5.5, 7 Important

Ex 5.5, 8

Ex 5.5, 9 Important

Ex 5.5, 10 Important

Ex 5.5, 11 Important

Ex 5.5, 12

Ex 5.5, 13

Ex 5.5, 14 Important

Ex 5.5, 15

Ex 5.5, 16 Important

Ex 5.5, 17 Important

Ex 5.5, 18

About the Author

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.