Ex 5.5, 8 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
Ex 5.5
Ex 5.5, 2
Ex 5.5, 3 Important
Ex 5.5, 4
Ex 5.5, 5
Ex 5.5,6 Important
Ex 5.5, 7 Important
Ex 5.5, 8 You are here
Ex 5.5, 9 Important
Ex 5.5, 10 Important
Ex 5.5, 11 Important
Ex 5.5, 12
Ex 5.5, 13
Ex 5.5, 14 Important
Ex 5.5, 15
Ex 5.5, 16 Important
Ex 5.5, 17 Important
Ex 5.5, 18
Last updated at April 16, 2024 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 ) )