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Ex 5.5

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Last updated at March 30, 2023 by Teachoo

Ex 5.5, 8 Differentiate the functions in, γ(sinβ‘π₯)γ^π₯+ sin^(β1) βπ₯ Let π¦=(sinβ‘π₯ )^π₯ + sin^(β1)β‘βπ₯ Let π’ = (sinβ‘π₯ )^π₯ & π£ = sin^(β1)β‘βπ₯ π¦ = π’ + π£ Differentiating both sides π€.π.π‘.π₯. ππ¦/ππ₯ = (π (π’ + π£))/ππ₯ ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Calculating π π/π π π’ =(sinβ‘π₯ )^π₯ Taking log both sides logβ‘π’=logβ‘γ(sinβ‘π₯ )^π₯ γ logβ‘π’=π₯ .logβ‘sinβ‘γπ₯ γ Differentiating both sides π€.π.π‘.π₯. (π(logβ‘π’))/ππ₯ = π(π₯.γ logγβ‘sinβ‘π₯ )/ππ₯ π(logβ‘π’ )/ππ₯ . ππ’/ππ’ = π(π₯ .γ logγβ‘sinβ‘π₯ )/ππ₯ π(logβ‘π’ )/ππ’ . ππ’/ππ₯ = π(π₯ .γ logγβ‘sinβ‘π₯ )/ππ₯ (As πππβ‘(π^π) = π . πππβ‘π) 1/π’ . ππ’/ππ₯ = π(π₯ .γ logγβ‘sinβ‘π₯ )/ππ₯ 1/π’ . ππ’/ππ₯ = ππ₯/ππ₯ . logβ‘sinβ‘π₯ + π(logβ‘sinβ‘π₯ )/ππ₯ Γ π₯ 1/π’ . ππ’/ππ₯ = 1 . logβ‘sinβ‘π₯ + (1/sinβ‘π₯ . π(sinβ‘π₯ )/ππ₯) π₯ 1/π’ . ππ’/ππ₯ = logβ‘sinβ‘π₯ + (1/sinβ‘π₯ . cosβ‘π₯ )π₯ 1/π’ . ππ’/ππ₯ = logβ‘sinβ‘π₯ + π₯ (cosβ‘π₯/sinβ‘π₯ ) ππ’/ππ₯ = u(logβ‘sinβ‘γπ₯+π₯ cotβ‘π₯ γ ) ππ’/ππ₯ = (sin π₯)^π₯ (logβ‘sinβ‘γπ₯+π₯ cotβ‘π₯ γ ) Using product Rule As (π’π£)β = π’βπ£ + π£βπ’ Calculating π π/π π π£=sin^(β1)β‘βπ₯ Differentiating both sides π€.π.π‘.π₯. ππ£/ππ₯ = π(sin^(β1)β‘βπ₯ )/ππ₯ ππ£/ππ₯ = 1/β(1 β(βπ₯)^2 ) . (πβπ₯)/ππ₯ ππ£/ππ₯ = 1/β(1 β π₯) . 1/(2βπ₯) ππ£/ππ₯ = 1/(2βπ₯ β(1 β π₯)) ππ£/ππ₯ = 1/(2β(π₯(1 βπ₯) ) ) ππ£/ππ₯ = 1/(2β(π₯ βπ₯^2 ) ) Now , ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Putting value of ππ’/ππ₯ & ππ£/ππ₯ π π/π π = (π¬π’π§β‘π )^π (π ππ¨πβ‘π+π₯π¨π β‘π¬π’π§β‘π ) + π/(π β(π β π^π ) )