Question

find the 2nd derivative of 2 sin x cos x

Explanation:

Step1: Simplify the function

Use the double - angle formula $\sin(2x)=2\sin x\cos x$. So the function $y = 2\sin x\cos x=\sin(2x)$.

Step2: Find the first - derivative

By the chain rule, if $y=\sin(u)$ and $u = 2x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Since $\frac{d}{du}\sin(u)=\cos(u)$ and $\frac{d}{dx}(2x)=2$, we have $y^\prime=\frac{d}{dx}\sin(2x)=2\cos(2x)$.

Step3: Find the second - derivative

Again, using the chain rule. Let $v = 2x$, and $y^\prime=2\cos(v)$. Then $\frac{d}{dx}(y^\prime)=\frac{d}{dv}(2\cos(v))\cdot\frac{dv}{dx}$. Since $\frac{d}{dv}\cos(v)=-\sin(v)$ and $\frac{dv}{dx}=2$, we get $y^{\prime\prime}=- 4\sin(2x)$.

Answer:

$-4\sin(2x)$