Answer
$-\cos x$
Work Step by Step
$\frac{d}{dx}(\sin x)=\cos x$
$\frac{d^{2}}{dx^{2}}(\sin x)= \frac{d}{dx}(\cos x)= -\sin x$
$\frac{d^{3}}{dx^{3}}(\sin x)=\frac{d}{dx}(-\sin x)= -\cos x$
$\frac{d^{4}}{dx^{4}}(\sin x)=\frac{d}{dx}(-\cos x)=\sin x$
This pattern will continue and we can generalise:
$\frac{d^{4n+1}}{dx^{4n+1}}(\sin x)=\cos x$
$\frac{d^{4n+2}}{dx^{4n+2}}(\sin x)=-\sin x$
$\frac{d^{4n+3}}{dx^{4n+3}}(\sin x)=-\cos x$
$\frac{d^{4n+4}}{dx^{4n+4}}(\sin x)=\sin x$
where n=0,1,2,3...
Therefore, $\frac{d^{99}}{dx^{99}}(\sin x)=\frac{d^{4\times24+3}}{dx^{4\times24+3}}(\sin x)= -\cos x$
