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Chapter 5 Class 12 Continuity and Differentiability
Ex 5.1, 13
Ex 5.1, 16
Ex 5.1, 18 Important
Ex 5.1, 28 Important
Ex 5.1, 30 Important
Ex 5.1, 34 Important
Ex 5.2, 5
Ex 5.2, 9 Important
Ex 5.2, 10 Important
Ex 5.3, 10 Important
Ex 5.3, 14 You are here
Example 29 Important
Example 30 Important
Ex 5.5,6 Important
Ex 5.5, 7 Important
Ex 5.5, 11 Important
Ex 5.5, 16 Important
Ex 5.6, 7 Important
Ex 5.6, 11 Important
Example 38 Important
Ex 5.7, 14 Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Example 39 (i)
Example 40 (i)
Example 42 Important
Misc 6 Important
Misc 15 Important
Misc 16 Important
Misc 22 Important
Chapter 5 Class 12 Continuity and Differentiability
Last updated at May 29, 2023 by Teachoo
Ex 5.3, 14 Find ππ¦/ππ₯ in, y = sinβ1 (2π₯ β(1βπ₯^2 )) , β 1/β2 < x < 1/β2 y = sinβ1 (2π₯ β(1βπ₯^2 )) Putting π₯ =π ππβ‘π π¦ = sinβ1 (2 sinβ‘π β(1βγπ ππγ^2 π)) π¦ = sinβ1 ( 2 sin ΞΈ β(γπππ γ^2 π)) π¦ ="sinβ1 " (γ"2 sin ΞΈ" γβ‘cosβ‘π ) π¦ = sinβ1 (sinβ‘γ2 π)γ π¦ = 2ΞΈ Putting value of ΞΈ = sinβ1 x π¦ = 2 γπ ππγ^(β1) π₯ Since x = sin ΞΈ β΄ γπ ππγ^(β1) x = ΞΈ Differentiating both sides π€.π.π‘.π₯ . (π(π¦))/ππ₯ = (π (γ2 sin^(β1)γβ‘π₯ ))/ππ₯ ππ¦/ππ₯ = 2 (πγ (π ππγ^(β1) π₯))/ππ₯ ππ¦/ππ₯ = 2 (1/β(1 βγ π₯γ^2 )) π π/π π = π/β(π β π^π ) ((sin^(β1)β‘π₯ )^β²= 1/β(1 β π₯^2 ))