Question

find the second derivative of 2x^{2}+sin 2x

Explanation:

Step1: Find first - derivative

Use the power rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$ and the chain - rule $\frac{d}{dx}(\sin(u))=\cos(u)\cdot\frac{du}{dx}$. Let $y = 2x^{2}+\sin(2x)$. Then $\frac{dy}{dx}=\frac{d}{dx}(2x^{2})+\frac{d}{dx}(\sin(2x))$.
$\frac{d}{dx}(2x^{2})=4x$ and $\frac{d}{dx}(\sin(2x))=\cos(2x)\cdot2 = 2\cos(2x)$. So, $\frac{dy}{dx}=4x + 2\cos(2x)$.

Step2: Find second - derivative

Differentiate $\frac{dy}{dx}=4x + 2\cos(2x)$ with respect to $x$. $\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(4x)+\frac{d}{dx}(2\cos(2x))$.
$\frac{d}{dx}(4x)=4$ and $\frac{d}{dx}(2\cos(2x))=2(-\sin(2x))\cdot2=-4\sin(2x)$.

Answer:

$4 - 4\sin(2x)$