Let $y = 3 \sin x + 2 \cos x$ Then $y\prime\prime$ is

A) $- 3 \sin x - 2 \cos x$
B) $- 3 \sin x + 2 \cos x$
C) $3 \sin x - 2 \cos x$
D) $3 \sin x + 2 \cos x$
E) $2 \sin x - 3 \cos x$

Question

Let $y = 3 \sin x + 2 \cos x$ Then $y\prime\prime$ is

A) $- 3 \sin x - 2 \cos x$
B) $- 3 \sin x + 2 \cos x$
C) $3 \sin x - 2 \cos x$
D) $3 \sin x + 2 \cos x$
E) $2 \sin x - 3 \cos x$

Quizplus · Accepted Answer

To find $y''$, we first differentiate $y$ with respect to $x$ to get $y'$, and then differentiate $y'$ again to get $y''$. The first derivative of $y = 3 \sin x + 2 \cos x$ is $y' = 3 \cos x - 2 \sin x$, using the derivatives $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$. Differentiating $y'$ again, $y'' = -3 \sin x - 2 \cos x$, using $\frac{d}{dx}(\cos x) = -\sin x$ and $\frac{d}{dx}(-\sin x) = -\cos x$.

Let y=3sin⁡x+2cos⁡xy = 3 \sin x + 2 \cos xy=3sinx+2cosx Then y′′y\prime\primey′′

Let $y = 3 \sin x + 2 \cos x$ Then $y\prime\prime$