Finding second order derivatives - Normal form

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

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Example 38 Find π2π¦/ππ₯2 , if π¦ = π₯3+tanβ‘π₯. π¦ = π₯3+tanβ‘π₯ Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = (π(π₯^3+ tanβ‘γπ₯)γ)/ππ₯ ππ¦/ππ₯ = (π(π₯^3))/ππ₯ + (π(tanβ‘γπ₯)γ)/ππ₯ ππ/ππ = πππ+ππππ π Again Differentiating π€.π.π‘.π₯ (π^2 π¦)/γππ₯γ^2 = (π (3π₯2 +sec^2β‘π₯))/ππ₯ (π^2 π¦)/γππ₯γ^2 = (π (3π₯2))/ππ₯ + (π (sec2 π₯))/ππ₯ (π^2 π¦)/γππ₯γ^2 = 6π₯+2 secβ‘π₯ . (π(secβ‘γπ₯)γ)/ππ₯ (π^2 π¦)/γππ₯γ^2 = 6π₯+2 secβ‘π₯.secβ‘γπ₯ tanβ‘π₯ γ γππγ^π/γππγ^π = ππ+π γπππγ^πβ‘π . π­ππ§β‘π