Question
Find the second-order derivatives of the function $x^2 + 3x + 2$

Answer

Let us take $y = x^2 + 3x + 2$
Now,
$\frac{d y}{d x}=\frac{d\left(x^{2}\right)}{d x}+\frac{d(3 x)}{d x}+\frac{d(2)}{d x}$
$= 2x + 3$
Therefore,
$\frac{d^{2} y}{d x^{2}}=\frac{d(2 x+3)}{d x}=\frac{d(2 x)}{d x}+\frac{d(3)}{d x}$
$= 2 + 0$
$= 2$

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