Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12



Last updated at March 11, 2021 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12
Transcript
Ex 5.5, 14 Find ππ¦/ππ₯ of the functions in, γ(cosβ‘γπ₯ γ)γ^π¦ = γ(cosβ‘γπ¦ γ)γ^π₯Given γ(cosβ‘π₯)γ^π¦ = γ(cosβ‘π¦)γ^π₯ Taking log both sides log γ(cosβ‘π₯)γ^π¦ = log γ(cosβ‘π¦)γ^π₯ π¦ . log (cosβ‘π₯)=π₯.logβ‘γ(cosβ‘π¦)γ Differentiating both sides π€.π.π‘.π₯. (π(π¦ . log (cosβ‘π₯)))/ππ₯ = π(π₯.γ logγβ‘γ(cosβ‘π¦)γ )/ππ₯ (As πππβ‘(π^π )=π . πππβ‘π) Finding (π (π . πππ (πππβ‘π)))/π π (π(π¦ . πππ (πππ β‘π₯)))/ππ₯ = (π(π¦))/ππ₯ . log coπ β‘π₯ + (π(πππ (πππ β‘π₯)))/ππ₯ . π¦ = ππ¦/ππ₯ . log coπ β‘π₯ + 1/πππ β‘π₯ . π(πππ β‘π₯ )/ππ₯ . π¦ = ππ¦/ππ₯ . log coπ β‘π₯ + 1/πππ β‘π₯ . (βsinβ‘π₯ ) . π¦ = ππ¦/ππ₯ . log coπ β‘π₯ + ((βsinβ‘π₯ ))/πππ β‘π₯ . π¦ = ππ¦/ππ₯ . log coπ β‘π₯βtanβ‘π₯. π¦ Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ Finding π (π.γ πππγβ‘γ(πππβ‘π)γ )/π π π(π₯.γ πππγβ‘γ(πππ β‘π¦)γ )/ππ₯ = (π(π₯))/ππ₯ . log coπ β‘π¦ + (π(πππ (πππ β‘π¦)))/ππ₯ . π₯ = log coπ β‘π¦ + 1/πππ β‘π¦ . π(πππ β‘π¦ )/ππ₯ . π₯ = log coπ β‘π¦ + 1/πππ β‘π¦ . π(πππ β‘π¦ )/ππ₯ . ππ¦/ππ¦ . π₯ = log coπ β‘π¦ + 1/πππ β‘π¦ . π(πππ β‘π¦ )/ππ¦ . ππ¦/ππ₯ . π₯ Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ = log coπ β‘π¦ + 1/πππ β‘π¦ . (βsinβ‘π¦) . ππ¦/ππ₯ . π₯ = log coπ β‘π¦ + βtanβ‘π¦ . π₯ . ππ¦/ππ₯ Now , (π(π¦ . log (cosβ‘π₯)))/ππ₯ = π(π₯.γ logγβ‘γ(cosβ‘π¦)γ )/ππ₯ ππ¦/ππ₯ log coπ β‘π₯βtanβ‘π₯. π¦ = log coπ β‘π¦ β tanβ‘π¦ . π₯ . ππ¦/ππ₯ ππ¦/ππ₯ log coπ β‘π₯βπ¦ . tanβ‘π₯ = log coπ β‘π¦ β π₯ . tanβ‘π¦ . ππ¦/ππ₯ ππ¦/ππ₯ log coπ β‘π₯+π₯ tan ππ¦/ππ₯ = log coπ β‘π¦ + π¦ tanβ‘π₯ ππ¦/ππ₯ (log coπ β‘π₯+π₯ tan π¦) = log coπ β‘π¦ + π¦ tanβ‘π₯ π π/π π = (π₯π¨π πππβ‘π " + " π πππβ‘π)/(π₯π¨π πππβ‘π + π πππ§ π)
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