Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 5.5

Ex 5.5, 1
Important

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 May 29, 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 ) )