Ex 5.5, 12 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at March 11, 2021 by
Last updated at March 11, 2021 by
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Ex 5.5, 12 Find ππ¦/ππ₯ of the functions in, π₯^π¦ + π¦^π₯ = 1 π₯^π¦ + π¦^π₯ = 1 Let π’ = π₯^π¦ , π£ = π¦^π₯ Hence, π’+π£=1 Differentiating both sides π€.π.π‘.π₯. (π(π£β‘γ+ π’γ))/ππ₯ = π(1)/ππ₯ ππ£/ππ₯ + ππ’/ππ₯ = 0 (Derivative of constant is 0) Calculating π π/π π π£=π₯^π¦ Taking log both sides logβ‘π£=logβ‘γ (π₯^π¦)" " γ logβ‘π£=γπ¦. logγβ‘π₯" " Differentiating both sides π€.π.π‘.π₯ (π(logβ‘π£))/ππ₯ = (π(π¦ . logβ‘π₯))/ππ₯ (π(logβ‘π£))/ππ₯ (ππ£/ππ£) = π(π¦ logβ‘π₯ )/ππ₯ (π(logβ‘π£))/ππ£ (ππ£/ππ₯) = π(γπ¦ logγβ‘π₯ )/ππ₯ 1/π£ (ππ£/ππ₯) = ( π(γπ¦ logγβ‘π₯ ))/ππ₯ 1/π£ (ππ£/ππ₯) = ( π(π¦))/ππ₯ . logβ‘π₯ + (π (logβ‘π₯))/ππ₯ . π¦ 1/π£ (ππ£/ππ₯) = ( ππ¦)/ππ₯ . logβ‘π₯ + 1/π₯ . π¦ 1/π£ (ππ£/ππ₯) = ( ππ¦)/ππ₯ logβ‘π₯ + π¦/π₯ ππ£/ππ₯ = v (log π₯+π¦/π₯) Putting value of π£ = π₯^π¦ ππ£/ππ₯ = π₯^π¦ (logβ‘γπ₯+ π¦/π₯γ ) ππ£/ππ₯ = π₯^π¦.log π₯ . ππ¦/ππ₯ + π₯^π¦. π¦/π₯ Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ Calculating π π/π π π’ = π¦^π₯ Taking log both sides logβ‘π’=logβ‘γ (π¦^π₯)" " γ logβ‘π’=γπ₯ . logγβ‘π¦" " Differentiating both sides π€.π.π‘.π₯ (π(logβ‘π’))/ππ₯ = (π(π₯ . logβ‘π¦))/ππ₯ (π(logβ‘π’))/ππ₯ (ππ’/ππ’) = π(π₯.logβ‘π¦ )/ππ₯ (π΄π logβ‘γ(π^π)γ=π logβ‘π ) (π(logβ‘π’))/ππ’ (ππ’/ππ₯) = (π (π₯ . logβ‘π¦))/ππ₯ 1/π’ . ππ’/ππ₯ = (π (π₯ . logβ‘π¦ ))/ππ₯ 1/π’ . ππ’/ππ₯ = (π (π₯))/ππ₯ . logβ‘π¦ + (π(logβ‘π¦))/ππ₯ . π₯ 1/π’ . ππ’/ππ₯ = 1 . logβ‘π¦ + π₯. π(logβ‘π¦ )/ππ₯ . ππ¦/ππ¦ 1/π’ . ππ’/ππ₯ = logβ‘π¦ + π₯. π(logβ‘π¦ )/ππ¦ . ππ¦/ππ₯ 1/π’ . ππ’/ππ₯ = logβ‘π¦ + π₯. 1/π¦ . ππ¦/ππ₯ Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ ππ’/ππ₯ = π’(logβ‘π¦ " + " π₯/π¦ " . " (ππ¦ )/ππ₯) ππ’/ππ₯ " = " π¦^π₯ (logβ‘π¦ " + " π₯/π¦ " . " (ππ¦ )/ππ₯) ππ’/ππ₯ = π¦^π₯ logβ‘π¦+π¦^(π₯ β1). π₯ ππ¦/ππ₯ Now, ππ£/ππ₯ + ππ’/ππ₯ = 0 Putting value of ππ£/ππ₯ & ππ’/ππ₯ (π₯^π¦ logβ‘γπ₯ γ ππ¦/ππ₯ +π₯^π¦ π¦/π₯) + (π¦^π₯ logβ‘γπ¦+π¦^(π₯ β1) γ.π₯ ππ¦/ππ₯) = 0 π₯^π¦ logβ‘γπ₯ γ ππ¦/ππ₯ +π₯^(π¦β1) π¦ + π¦^π₯ logβ‘γπ¦+π¦^(π₯ β1) γ.π₯ ππ¦/ππ₯ = 0 π₯^π¦ logβ‘γπ₯ γ ππ¦/ππ₯+π¦^(π₯ β1).π₯ ππ¦/ππ₯=β(π₯^(π¦β1) π¦+π¦^π₯ logβ‘π¦ ) ππ¦/ππ₯(π₯^π¦ logβ‘γπ₯ γ+π¦^(π₯ β1).π₯)=β(π₯^(π¦β1) π¦+π¦^π₯ logβ‘π¦ ) π π/π π = (β(π^(πβπ) π + π^π πππβ‘π ))/((π^π πππβ‘γπ γ+ π^(π βπ).π))
Logarithmic Differentiation - Type 2
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