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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Example 39 (i) - Chapter 5 Class 12 Continuity and Differentiability
Last updated at June 2, 2023 by Teachoo
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Example 39 (i) You are here
Example 39 (ii) Important
Example 40 (i)
Example 40 (ii) Important
Example 40 (iii) Important
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Question 1 Deleted for CBSE Board 2024 Exams
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Transcript
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))
