Найдите дельта х и дельта f в точке х0 1)f(x)=pi/3+sin x; x0=2pi/3; x=3pi/4 2)f(x)=ctg x...

1) $x_{0}=\frac{2\pi}{3}\\ x=(x_{0}+\Delta x)=\frac{3\pi}{4}\\ \Delta x=x-x_{0}=\frac{3\pi}{4}-\frac{2\pi}{3}=\frac{9\pi-8\pi}{12}=\frac{\pi}{12}\\ \Delta y=f(x_{0}+\Delta x)-f(x_{0})\\ \Delta y=(\frac{\pi}{3}+sin\frac{3\pi}{4})-(\frac{\pi}{3}+sin\frac{2\pi}{3})=sin\frac{3\pi}{4}-sin\frac{2\pi}{3}=\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}=\\ =\frac{\sqrt{2}-\sqrt{3}}{2}$

2) $x_{0}=\frac{\pi}{4}\\ x=(x_{0} +\Delta x) = \frac{\pi}{3}\\ \Delta x = x-x_{0}=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}\\ \Delta y=f(x_{0}+\Delta x) -f(x_{0})\\ \Delta y =(ctg\frac{\pi}{3}-\frac{\pi}{4})-(ctg\frac{\pi}{4}-\frac{\pi}{4})=ctg\frac{\pi}{3}-ctg\frac{\pi}{4}=\frac{1}{\sqrt{3}} -1=\frac{\sqrt{3}-3}{3}$