Question

Find the second order derivatives of $x.\cos x$

Answer 1

Let $y=x.\cos x$
$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(x.\cos x)=\cos x\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(\cos x\right)=\cos x.1+x(-\sin x)=\cos x-x\sin x$
$\therefore$ $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}[\cos x-x\sin x]=\frac{d}{dx}\left(\cos x\right)-\frac{d}{dx}\left(x\sin x\right)$
$=-\sin x-\left[\sin x\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(\sin x\right)\right]$
$=-\sin x-(\sin x+x\cos x)$
$=-(x\cos x+2\sin x)$