Question

Find the second order derivatives of the function
$x.cosx$

Answer 1

The correct answer is $=-(xcosx+2sinx)$
Let $y=x.cosx$
Then,
$\frac{dy}{dx}=\frac{d}{dx}(x.cosx)=cosx.\frac{d}{dx}(x)+x\frac{d}{dx}(cosx)$
$=cosx.1+x(-sinx)=cosx-xsinx$
$∴\frac{d^2y}{dx^2}=\frac{d}{dx}(cosx-xsinx)=\frac{d}{dx}(cosx)-\frac{d}{dx}(xsinx)$
$=-sinx-[sinx.\frac{d}{dx}(x)+x.\frac{d}{dx}(sinx)]$
$=-sinx-(sinx+xcosx)$
$=-(xcosx+2sinx)$