Answer
$-cos x$
Work Step by Step
$\frac{d}{dx}[sin x]=cos x $
$\frac{d^{2}}{d^2x}[sin x]=-sin x $
$\frac{d^{3}}{d^3x}[sin x]=-cos x $
$\frac{d^{4}}{d^4x}[sin x]=sin x $
Here we return to the original function, so the pattern will repeat each 4 derivatives. For example, the multiples of 4 will be just like the 4th derivative: 8, 12, 16,... And the ones before them will be just like the 3rd derivative: 7, 11, 15...
We want the 87th derivative, 88 is a multiple of 4, since 4•22=88, and 87 is before 88, which means:
$\frac{d^{87}}{d^{87}x}[sin x]=\frac{d^{3}}{d^3x}[sin x]=-cos x $
