Ex 5.7, 6 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
Last updated at April 16, 2024 by Teachoo
Ex 5.7, 6 Find the second order derivatives of the function π^π₯ sinβ‘5π₯ Let y = π^π₯ sinβ‘5π₯ Differentiating π€.π.π‘.π₯ . ππ¦/ππ₯ = (π(π^π₯ " " sinβ‘5π₯))/ππ₯ ππ¦/ππ₯ = π(π^π₯ )/ππ₯ .sinβ‘γ 5π₯γ + (π(γsin 5γβ‘π₯))/ππ₯ . π^π₯ ππ¦/ππ₯ =π^π₯ .sinβ‘γ 5π₯γ + cosβ‘5π₯ (π(5π₯))/ππ₯ . π^π₯ ππ¦/ππ₯ = π^π₯. sinβ‘γ 5π₯γ + 5.π^π₯. cosβ‘5π₯ using product rule in π^π₯ π ππβ‘5π₯ As (π’π£)β= π’βπ£ + π£βπ’ Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯)= (π (π^π₯. sinβ‘γ 5π₯γ " + " 5.π^π₯." " cosβ‘5π₯))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = (π (π^π₯. sinβ‘γ 5π₯γ))/ππ₯ + (π (5.π^π₯." " cosβ‘5π₯))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = (π (π^π₯ sinβ‘5π₯))/ππ₯ + 5 (π(π^π₯." " cosβ‘5π₯))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = ((π (π^π₯))/ππ₯.sinβ‘5π₯+(π (sinβ‘5π₯))/ππ₯ .π^π₯ ) + 5 ((π(π^π₯))/ππ₯ .cosβ‘5π₯+(π(cosβ‘5π₯))/ππ₯ .π^π₯ ) using product rule in π^π₯ π ππβ‘5π₯ & π^π₯." " πππ β‘5π₯ As (π’π£)β= π’βπ£ + π£βπ’ (π^2 π¦)/(ππ₯^2 ) = (π^π₯.sinβ‘5π₯+(cosβ‘5π₯ ) .(π(β‘5π₯))/ππ₯.π^π₯ ) + 5 (π^π₯.cosβ‘5π₯+(βsinβ‘γ5 π₯)γ.(π(β‘5π₯))/ππ₯.π^π₯ ) (π^2 π¦)/(ππ₯^2 ) = (π^π₯ sinβ‘5π₯+cosβ‘5π₯.5.π^π₯ ) + 5 (π^π₯ cosβ‘5π₯βsinβ‘γ5 π₯γ.5.π^π₯ ) (π^2 π¦)/(ππ₯^2 ) = π^π₯ sinβ‘5π₯+5π^π₯ cosβ‘5π₯+5π^π₯ cosβ‘5π₯β25π^π₯ sinβ‘γ5 π₯γ (π^2 π¦)/(ππ₯^2 ) = 10 π^π₯ cosβ‘5π₯ β 24 π^π₯ sinβ‘γ5 π₯γ (π ^π π)/(π π^π ) = 2π^π (ππππβ‘ππ β 1π πππβ‘γπ πγ)