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

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.