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Example 44 (i) - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at Aug. 19, 2021 by Teachoo
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Example 29 (i)
Example 29 (ii) Important
Example 29 (iii) Important
Example 29 (iv)
Example 30 Important
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Example 33 Important
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Example 37 Important
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Example 39 Important
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Example 42 Important Deleted for CBSE Board 2022 Exams
Example 43 Deleted for CBSE Board 2022 Exams
Example 44 (i) You are here
Example 44 (ii) Important
Example 44 (iii) Important
Example 45 (i)
Example 45 (ii) Important
Example 45 (iii) Important
Example 46
Example 47 Important
Example 48
Transcript
Example 44 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))
