Finding second order derivatives - Normal form

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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### Transcript

Ex 5.7, 5 Find the second order derivatives of the function π₯^3 logβ‘π₯ Let y = π₯^3 logβ‘π₯ Differentiating π€.π.π‘.π₯ . ππ¦/ππ₯ = (π(π₯^3 " " logβ‘π₯))/ππ₯ ππ¦/ππ₯ = π(π₯^3 )/ππ₯ .logβ‘π₯ + (π(logβ‘π₯))/ππ₯ .π₯^3 using product rule in π₯^3 πππβ‘π₯ . As (π’π£)β= π’βπ£ + π£βπ’ where u = x3 & v = log x ππ¦/ππ₯ = 3π₯2 . logβ‘π₯ + 1/π₯ . π₯^3 ππ¦/ππ₯ = 3π₯2 . logβ‘π₯ + π₯^2 Again Differentiating π€.π.π‘.π₯ π/ππ₯ (ππ¦/ππ₯) = (π (3π₯2 . logβ‘π₯ "+ " π₯^2))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = (π (3π₯2 . logβ‘π₯))/ππ₯ + (π (π₯^2))/ππ₯ (π^2 π¦)/(ππ₯^2 ) = (3 π(π₯2 . logβ‘π₯))/ππ₯ + 2 π₯ using product rule in π₯^2 πππβ‘π₯ . As (π’π£)β= π’βπ£ + π£βπ’ (π^2 π¦)/(ππ₯^2 ) = 3 . ((π(π₯)^2)/ππ₯ .logβ‘γπ₯+(π(logβ‘γπ₯)γ)/ππ₯γ. π₯^2 ) + 2 π₯ = 3(2π₯ . logβ‘γπ₯+ 1/π₯γ. π₯^2 ) + 2 π₯ = 3 (2π₯ . logβ‘π₯+ π₯) + 2 π₯ = 6 π₯ logβ‘π₯ + 3π₯ + 2π₯ = 6 π₯ log β‘π₯+5π₯ = x (6 logβ‘γπ₯+5γ ) = π₯ (5+6 logβ‘π₯ ) Hence , (π^π π)/(ππ^π ) = π (π+π πππβ‘π )