Find $∆y$ and $dy$ for the function $y= \cos x$ at $\mathrm{x}={60}^0$ with $\mathrm{\Delta }x=1^0$ where $\cos {61}^0=0.4848,1^0=0.0174$

$f\left(x\right)=\cos x$

$\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f\left(x\right)=\cos (x+\mathrm{\Delta }x)-\cos x$

$=\cos \left({60}^0+1^0\right)-\cos {60}^{0}s=\cos {61}^0-\cos {60}^{\circ }$

$=0.4848-0.5=-0.0152$

$dy=f^1\left(x\right)\cdot \mathrm{\Delta }x=-\sin x\cdot \left(1^0\right)=-\sin 60.\left(1^0\right)$

$=-0.866\times 0.0174=-0.015$