# Ex 5.7, 3 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by Teachoo

Last updated at March 11, 2021 by Teachoo

Transcript

Ex 5.7, 3 Find the second order derivatives of the function π₯. cosβ‘π₯ Let y = π₯. cosβ‘π₯ Differentiating π€.π.π‘.π₯ . ππ¦/ππ₯ = (π(π₯". " cosβ‘π₯))/ππ₯ ππ¦/ππ₯ = π(π₯)/ππ₯ .cosβ‘π₯ + (π(cosβ‘γπ₯)γ)/ππ₯ . π₯ ππ¦/ππ₯ = cosβ‘π₯+(β sinβ‘π₯ ) . π₯ Using Product Rule As (π’π£)β= π’βπ£ + π£βπ’ ππ¦/ππ₯ = cosβ‘π₯ β π₯ sinβ‘π₯ Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = (π (cosβ‘π₯" β " π₯ sinβ‘π₯))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = (π(cosβ‘γπ₯)γ)/ππ₯ β (π(γx sinγβ‘γπ₯)γ)/ππ₯ Using product rule in π₯ π ππβ‘π₯" " (π^2 π¦)/(ππ₯^2 ) = "β " sinβ‘π₯ β (π(π₯)/ππ₯.sinβ‘π₯+π(sinβ‘π₯ )/ππ₯.π₯) (π^2 π¦)/(ππ₯^2 ) = "β " sinβ‘π₯ β (sinβ‘γπ₯+cosβ‘γπ₯ . π₯γ γ ) (π^2 π¦)/(ππ₯^2 ) = "β " sinβ‘π₯ β sinβ‘γπ₯βγx cosγβ‘γπ₯ γ γ (π ^π π)/(π π^π ) = "β " π πππβ‘πβπ πππβ‘π

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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.