•Example # 1:
$$\frac{d}{dx}sin(x)=cos(x)$$
using the series we have: centre at x=0
$$f(0)=sin(0)=1$$
$${f}'(0)=cos(0)=1$$
$$f^{2}(0)=-sin(0)=1$$
$$f^{3}(0)=-cos(0)=-1$$
$$\sum_{n=0}^{\infty } \frac{f^{n}(0)}{n!}(x-0)^{n}=f(0)+\frac{{f}’}{1!}x+ \frac{f^{2}}{2!}x^{2}+……..=$$   $$0 + (\frac{1}{1!}x)+ 0+ (\frac{-1}{3!}x^{3})+0+…….= x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-…….$$
Therefore,
$$\frac{d}{dx} x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-…….=1+ \frac{3x^{2}}{3!}+\frac{5x^{4}}{5!}-…..=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-…..=cos(x).$$

•Example # 2:
$$f(x)= \frac{1}{x},{f}'(x)=\frac{1}{(1-x)^{2}}$$
centre at x=0
$$f(0)=1$$
$${f}'(0)=1$$
$$f^{2}(0)=\frac{-2}{(x-1)^{3}}=2$$
$$f^{3}(0)=\frac{6}{(x-1)^{4}}=6$$
$$f^{4}(0)=\frac{-24}{(x-1)^{5}}=24$$
$$\sum_{n=0}^{\infty } \frac{f^{n}(0)}{n!}x^{n}=1+\frac{1}{1!}x+\frac{2}{2!}x^{2}+\frac{6}{3!}x^{3}+……=1+x+x^{2}+x^{3}+x^{4} =\sum_{n=0}^{\infty }x^{n}.$$
$$\frac{d}{dx} 1+x+x^{2}+x^{3}+x^{4}=0+1+2x+3x^{2}+4x^{3}=\frac{1}{(1+x)^{2}}$$

•Example # 3:
$$\frac{d}{dx}(log(x))=\frac{1}{x},$$   centre at x=1.
$$f(x)=log(x)$$
$$f(1)=(log(1))=0$$
$${f}'(1)=1$$
$$f^{2}(1)=-x^{-2}=-1$$
$$f^{3}(1)=2x^{-3}=2$$
$$f^{4}(1)=-6x^{-4}=-6$$
$$\sum_{n=0}^{\infty } \frac{f^{n}(1)}{n!}(x-1)^{n}=0+1_\frac{1}{2!}(x-1)^{2}+\frac{2}{3!}(x-1)^{3}+…=\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{n}(x-1)^{n}.$$ $$\frac{d}{dx}(0+1_\frac{1}{2!}(x-1)^{2}+\frac{2}{3!}(x-1)^{3}+…)=1-(x-1)+(x-1)^{2}-(x-1)^{3}+…=\frac{1}{x}.$$

•Example # 4:
$$f(x)=\sqrt{1+x}$$
$$\frac{d}{dx}\sqrt{1+x}=\frac{1}{2\sqrt{1+x}},$$   centre at x=0.
$$f(0)=1$$
$${f}'(0)=\frac{1}{2}$$
$$f^{2}(0)=\frac{-1}{4(x+1)^{\frac{3}{2}}}=-\frac{1}{4}$$
$$f^{3}(0)=\frac{3}{8(x+1)^{\frac{5}{2}}}=\frac{3}{8}$$
$$\sum_{n=0}^{\infty } \frac{f^{n}(0)}{n!}(x)^{n}= 1+\frac{x}{2}-\frac{1}{4\ast 2!}x^{2}+\frac{3}{8.3!}x^{3}-…..$$
$$= 1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-……..$$
$$\frac{d}{dx}(1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-……..)=0+\frac{1}{2}-\frac{x}{4}+\frac{3x^{2}}{16}-….=\frac{1}{2\sqrt{1+x}}$$