Finding second order derivatives - Normal form

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month

Transcript

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₯ . ππ¦/ππ₯ = (π(π^6π₯ " " cosβ‘3π₯))/ππ₯ ππ¦/ππ₯ = π(π^6π₯ )/ππ₯ .cosβ‘3π₯ + (π(cosβ‘3π₯))/ππ₯ .γ πγ^6π₯ ππ¦/ππ₯ =γ πγ^6π₯ .π(6π₯)/ππ₯ . cosβ‘3π₯ + (γβsinγβ‘3π₯) . (π(3π₯))/ππ₯ . γ πγ^6π₯ Using product rule in π^6π₯ πππ β‘3π₯ . As (π’π£)β= π’βπ£ + π£βπ’ where u = π^6π₯ " & v =" cosβ‘3π₯ ππ¦/ππ₯ = π^6π₯ . 6 . cosβ‘3π₯ β sinβ‘3π₯ . 3 . π^6π₯ ππ¦/ππ₯ = 3γ πγ^6π₯ (2γ cosγβ‘3π₯ β sin 3π₯) Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = (π(3 π^6π₯ (2 cosβ‘3π₯ β sinβ‘3π₯) ))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = 3(π(π^6π₯ (2 cosβ‘3π₯ β sinβ‘3π₯) ))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = 3 (π(γ πγ^6π₯ )/ππ₯ ."(2" cosβ‘3π₯ " β sin " 3π₯") " +(π(2 cosβ‘3π₯ β sinβ‘3π₯))/ππ₯ ". " π^6π₯ ) Using product Rule As (π’π£)β= π’βπ£ + π£βπ’ where v = γ πγ^6π₯ & v = 2 cos 3x β sin 3x (π^2 π¦)/(ππ₯^2 ) = 3 [6γ πγ^6π₯ "(2" γ cosγβ‘3π₯ " β sin " 3π₯") + (" β"2" sinβ‘γ3π₯.3γ " β " cosβ‘3π₯.3")" π^6π₯ ] (π^2 π¦)/(ππ₯^2 ) = 3[12γ πγ^6π₯ γ.cosγβ‘3π₯ "β 6" γ πγ^6π₯ "sin " 3π₯βγ6 πγ^6π₯ " sin " 3π₯β3γ πγ^6π₯ cosβ‘3π₯ ] (π^2 π¦)/(ππ₯^2 ) = 3[9π^6π₯ cosβ‘3π₯β12π^6π₯.sinβ‘3π₯ ] (π^2 π¦)/(ππ₯^2 ) = 9π^6π₯ [3 cosβ‘3π₯β4 sinβ‘3π₯ ] Hence , (π^π π)/(ππ^π ) = ππ^ππ [π πππβ‘ππβπ πππβ‘ππ ]

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.