Ex 5.3, 1 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at March 10, 2021 by Teachoo
Transcript
Ex 5.3, 1 Find ππ¦/ππ₯ in, 2π₯ + 3π¦ = sinβ‘π₯ Given 2π₯+3π¦ = sinβ‘π₯ Differentiating both sides π€.π.π‘.π₯ (π (2π₯ + 3π¦))/(ππ₯ ) = (π(sinβ‘γπ₯)γ)/(ππ₯ ) (π(2π₯))/(ππ₯ ) +(π(3π¦))/(ππ₯ )=(π(sinβ‘γπ₯)γ)/(ππ₯ ) 2 (π(π₯))/(ππ₯ ) +"3" (π(π¦))/(ππ₯ )=(π(sinβ‘γπ₯)γ)/(ππ₯ ) 2 + 3 (π(π¦))/(ππ₯ ) = cosβ‘π₯ 3 (π(π¦))/(ππ₯ ) = cosβ‘π₯ β 2 (π (π))/(π π ) = π/π (πππβ‘π β 2) (Derivative of π ππβ‘π₯ is πππ β‘π₯)
About the Author
Davneet Singh
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.