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Example 47 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at March 22, 2023 by Teachoo
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Transcript
Example 47 For a positive constant a find ππ¦/ππ₯ , where π¦ = π^(π‘+1/π‘) , and π₯ =(π‘+1/π‘)^2 Here π π/π π = (π π/π π)/(π π/π π) Calculating π π/π π π¦=π^(π‘ + 1/π‘) Differentiating π€.π.π‘. t π π/π π = π (π^((π + π/π) ) )/π π ππ¦/ππ‘ = π^((π‘ + 1/π‘) ) .logβ‘π.π(π‘ + 1/π‘)/ππ‘ ππ¦/ππ‘ = π^((π‘ + 1/π‘) ) .logβ‘π.(1+(β1) π‘^(β2) ) π π/π π = π^((π + π/π) ) .πππβ‘π.(πβπ/π^π ) "As " π(π^π₯ )/ππ₯ " = " π^π₯.πππβ‘π Calculating π π/π π π₯=(π‘+1/π‘)^π Differentiating π€.π.π‘. t ππ₯/ππ‘ = π((π‘ + 1/π‘)^(π ) )/ππ‘ ππ₯/ππ‘ = a (π‘+1/π‘)^(π β1 ) . π(π‘ + 1/π‘)/ππ‘ ππ₯/ππ‘ = a (π‘+1/π‘)^(π β1 ) . (π(π‘)/ππ‘ + π(1/π‘)/ππ‘) ππ₯/ππ‘ = a (π‘+1/π‘)^(π β1 ) . (1+ π(π‘^(β1) )/ππ‘) ππ₯/ππ‘ = a π^(π β1 ) . π(π)/ππ‘ ππ₯/ππ‘ = a (π‘+1/π‘)^(π β1 ) . (1+(β1) γ π‘γ^(β2) ) ππ₯/ππ‘ = a (π‘+1/π‘)^(π β1 ) . (1β 1/π‘^2 ) Calculating π π/π π ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = (π^(π‘ + 1/π‘) . logβ‘γπ γ Γ (1 β 1/π‘^2 ))/(π(π‘ + 1/π‘)^(π β 1) (1 β 1/π‘^2 ).) π π/π π = (π^(π + π/π) . πππβ‘γπ γ)/(π(π + π/π)^(π β π) )
