# Ex 5.7, 7 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by

Last updated at March 11, 2021 by

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Ex 5.7, 7 Find the second order derivatives of the function π^6π₯ cosβ‘3π₯ Let y = π^6π₯ cosβ‘3π₯ Differentiating π€.π.π‘.π₯ . ππ¦/ππ₯ = (π(π^6π₯ " " cosβ‘3π₯))/ππ₯ ππ¦/ππ₯ = π(π^6π₯ )/ππ₯ .cosβ‘3π₯ + (π(cosβ‘3π₯))/ππ₯ .γ πγ^6π₯ ππ¦/ππ₯ =γ πγ^6π₯ .π(6π₯)/ππ₯ . cosβ‘3π₯ + (γβsinγβ‘3π₯) . (π(3π₯))/ππ₯ . γ πγ^6π₯ Using product rule in π^6π₯ πππ β‘3π₯ . As (π’π£)β= π’βπ£ + π£βπ’ where u = π^6π₯ " & v =" cosβ‘3π₯ ππ¦/ππ₯ = π^6π₯ . 6 . cosβ‘3π₯ β sinβ‘3π₯ . 3 . π^6π₯ ππ¦/ππ₯ = 3γ πγ^6π₯ (2γ cosγβ‘3π₯ β sin 3π₯) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = (π(3 π^6π₯ (2 cosβ‘3π₯ β sinβ‘3π₯) ))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = 3(π(π^6π₯ (2 cosβ‘3π₯ β sinβ‘3π₯) ))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = 3 (π(γ πγ^6π₯ )/ππ₯ ."(2" cosβ‘3π₯ " β sin " 3π₯") " +(π(2 cosβ‘3π₯ β sinβ‘3π₯))/ππ₯ ". " π^6π₯ ) Using product Rule As (π’π£)β= π’βπ£ + π£βπ’ where v = γ πγ^6π₯ & v = 2 cos 3x β sin 3x (π^2 π¦)/(ππ₯^2 ) = 3 [6γ πγ^6π₯ "(2" γ cosγβ‘3π₯ " β sin " 3π₯") + (" β"2" sinβ‘γ3π₯.3γ " β " cosβ‘3π₯.3")" π^6π₯ ] (π^2 π¦)/(ππ₯^2 ) = 3[12γ πγ^6π₯ γ.cosγβ‘3π₯ "β 6" γ πγ^6π₯ "sin " 3π₯βγ6 πγ^6π₯ " sin " 3π₯β3γ πγ^6π₯ cosβ‘3π₯ ] (π^2 π¦)/(ππ₯^2 ) = 3[9π^6π₯ cosβ‘3π₯β12π^6π₯.sinβ‘3π₯ ] (π^2 π¦)/(ππ₯^2 ) = 9π^6π₯ [3 cosβ‘3π₯β4 sinβ‘3π₯ ] Hence , (π ^π π)/(π π^π ) = ππ^ππ [π πππβ‘ππβπ πππβ‘ππ ]

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.