Miscellaneous
Last updated at Dec. 16, 2024 by Teachoo
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Misc 9 Differentiate w.r.t. x the function, (sinβ‘π₯βcosβ‘π₯ )^((sinβ‘γπ₯βcosβ‘γπ₯)γ γ ), π/4 <π₯< 3π/4 Let y = (sinβ‘π₯βcosβ‘π₯ )^((sinβ‘γπ₯βcosβ‘γπ₯)γ γ ) Taking log on both sides logβ‘π¦ = log (sinβ‘π₯βcosβ‘π₯ )^((sinβ‘γπ₯βcosβ‘γπ₯)γ γ ) logβ‘π¦ = (sinβ‘π₯βcosβ‘π₯ ). γ logγβ‘γ (sinβ‘π₯βcosβ‘π₯ )γ Differentiating both sides π€.π.π‘.π₯. π(logβ‘π¦ )/ππ₯ = π((sinβ‘π₯ β cosβ‘π₯ ). γ logγβ‘(sinβ‘π₯ β cosβ‘π₯ ) )/ππ₯ π(logβ‘π¦ )/ππ₯ (ππ¦/ππ¦) = π((sinβ‘π₯βcosβ‘π₯ ). γ logγβ‘γ (sinβ‘π₯βcosβ‘π₯ )γ )/ππ₯ π(logβ‘π¦ )/ππ¦ (ππ¦/ππ₯) = π((sinβ‘π₯βcosβ‘π₯ ). γ logγβ‘γ (sinβ‘π₯βcosβ‘π₯ )γ )/ππ₯ 1/π¦ . ππ¦/ππ₯ = π((sinβ‘π₯βcosβ‘π₯ ). γ logγβ‘γ (sinβ‘π₯βcosβ‘π₯ )γ " " )/ππ₯ " " 1/π¦. ππ¦/ππ₯ = π(sinβ‘π₯ β cosβ‘π₯ )/ππ₯ . γ log γβ‘(sinβ‘π₯βcosβ‘π₯ ) + π(γ logγβ‘γ (sinβ‘π₯ β cosβ‘π₯ )γ )/ππ₯ .(sinβ‘π₯βcosβ‘π₯ ) 1/π¦ . ππ¦/ππ₯ = (cosβ‘π₯β(βsinβ‘π₯ )). logβ‘γ (sinβ‘π₯βγ cosγβ‘π₯ )γ + 1/((sinβ‘π₯ β cosβ‘π₯ ) ) . π(sinβ‘π₯ β cosβ‘π₯ )/ππ₯ . (sinβ‘π₯βγ cosγβ‘π₯ ) 1/π¦ . ππ¦/ππ₯ = (cosβ‘π₯+sinβ‘π₯ ) . logβ‘γ (sinβ‘π₯βcosβ‘π₯ )γ + 1/((sinβ‘π₯βcosβ‘π₯ ) ) . (cosβ‘π₯β(βsinβ‘π₯ )) . (sinβ‘π₯βcosβ‘π₯ ) Using product rule (π’π£)β = π’βπ£ + π£βπ’ where u = sin x β cos x & v = log (sin x β cos x) 1/π¦ . ππ¦/ππ₯ = (cosβ‘π₯+sinβ‘π₯ ) . logβ‘γ (sinβ‘π₯βcosβ‘π₯ )γ + 1/((sinβ‘π₯βcosβ‘π₯ ) ) . (cosβ‘π₯+sinβ‘π₯ ) . (sinβ‘π₯βcosβ‘π₯ ) 1/π¦ . ππ¦/ππ₯ = (cosβ‘π₯+sinβ‘π₯ ) . logβ‘γ (sinβ‘π₯βcosβ‘π₯ )γ + (cosβ‘π₯+sinβ‘π₯ ) 1/π¦ . ππ¦/ππ₯ = (cosβ‘π₯+sinβ‘π₯ ) . (logβ‘γ (sinβ‘π₯βcosβ‘π₯ )+1γ ) ππ¦/ππ₯ = π¦(cosβ‘π₯+sinβ‘π₯ ) . (logβ‘γ (sinβ‘π₯βcosβ‘π₯ )+1γ ) π π/π π = (πππβ‘πβπππβ‘π )^((πππβ‘γπβπππβ‘γπ)γ γ ) (πππβ‘π+πππβ‘π )(πππβ‘γ (πππβ‘πβπππβ‘π )+πγ )