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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
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) You are here
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 June 2, 2023 by Teachoo
Example 39 Differentiate w.r.t. x, the following function: (i) β(3π₯+2) + 1/β(2π₯^2+ 4) Let y = β(3π₯+2) + 1/β(2π₯^2+ 4 ) Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π(β(3π₯ + 2) " + " 1/β(2π₯^2 + 4 ))/ππ₯ ππ¦/ππ₯ = π(β(3π₯ + 2))/ππ₯ + π(1/β(2π₯^2 + 4 ))/ππ₯ ππ¦/ππ₯ = π(β(3π₯ + 2))/ππ₯ + (π(2π₯^2 + 4)^((β1)/2))/ππ₯ Calculating π(β(3π₯ + 2))/ππ₯ & (π(2π₯^2 + 4)^((β1)/2))/ππ₯ separately Calculating π(β(ππ± + π))/π π π(β(3π₯ + 2))/ππ₯ = 1/(2β(3π₯ + 2 )) Γ π(3π₯ + 2)/ππ₯ = 1/(2β(3π₯ + 2 )) Γ (3+0) = π/(πβ(ππ + π )) Calculating (π (ππ^π + π)^((βπ)/π))/π π (π(2π₯^2 + 4)^((β1)/2))/ππ₯ = (β1)/2 γ(2π₯^2+4)γ^((β1)/( 2) β1) . π(2π₯^2+ 4)/ππ₯ = (β1)/2 (2π₯^2+ 4)^((β3)/( 2)) . (π(2π₯^2 )/ππ₯ + π(4)/ππ₯) = (β1)/2 (2π₯^2+ 4)^((β3)/( 2)) . (4π₯+0) = (β4π₯)/2 (2π₯^2+ 4)^((β3)/( 2)) = (βππ)/(ππ^π+ π)^(π/π) Hence, ππ¦/ππ₯ = π(β(3π₯+2))/ππ₯ + π(1/β(2π₯^2+ 4 ))/ππ₯ π π/π π = π/(πβ(ππ + π )) β ππ/(ππ^π+ π)^(π/( π))