Ex 5.7, 6 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 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)
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo