Question

find y
y = 3x cos x
y = □

Explanation:

Step1: Apply product - rule for first - derivative

The product rule is $(uv)' = u'v+uv'$, where $u = 3x$ and $v=\cos x$. So $y'=(3x)'\cos x+3x(\cos x)'$. Since $(3x)' = 3$ and $(\cos x)'=-\sin x$, we have $y'=3\cos x - 3x\sin x$.

Step2: Apply product - rule for second - derivative

We need to find the derivative of $y' = 3\cos x-3x\sin x$. The derivative of $3\cos x$ is $- 3\sin x$. For the second term $-3x\sin x$, using the product rule with $u=-3x$ and $v = \sin x$, we have $(-3x\sin x)'=(-3x)'\sin x+(-3x)(\sin x)'=-3\sin x-3x\cos x$. Then $y''=-3\sin x-3\sin x - 3x\cos x=-6\sin x-3x\cos x$.

Answer:

$-6\sin x - 3x\cos x$