Ex 5.3, 1 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at Dec. 7, 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 ππ¦/(ππ₯ ) = πππβ‘π
3 ππ¦/(ππ₯ ) = cosβ‘π₯ β 2
π
π/(π
π ) = π/π (πππβ‘π β 2)

Show More