Question
Find $\frac{d^{2} y}{d x^{2}}$, if $y = x^3$ + tan x.

Answer

Given that $y = x^3 + \tan x$. Then
$\frac{d y}{d x}=3x^2 + \sec^2 x$
$\therefore$ $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(3 x^{2}+\sec ^{2} x\right)$
$= 6x + 2 sec x. sec\ x\ tan\ x$
$= 6x + 2 \sec^2 x \tan x$

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