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\sum\limits_{i=1}^n{\frac{1}{(i)(i+1)}} =\sum\limits_{i=1}^n{\frac{1}{(i)}-\frac{1}{(i+1)}} =\sum\limits_{i=1}^n{\frac{1}{i}} -\sum\limits_{i=1}^{n}\frac{1}{i+1} \\=(\frac{1}{1}+\sum\limits_{i=2}^{n}{\frac{1}{i}}) -(\sum\limits_{i=2}^{n}\frac{1}{i}+\frac{1}{n+1}) \\=1-\frac{1}{1+n} \\ 令:
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f(x)=\frac{2x-1}{x+1}=\frac{2x-1+2-2}{\frac{1}{2}(2x+2)} =\frac{2x+2-3}{x+1}=2+\frac{-3}{x+1}
f(x)=x+12x−1=21(2x+2)2x−1+2−2=x+12x+2−3=2+x+1−3
再比如导函数求导法则种的乘法求导法则的推导/证明:
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f(x)=u(x)v(x) \\ f'(x)=\lim_{\Delta x\rightarrow 0} {\frac{f(x+\Delta x)-f(x)}{\Delta x}} \\=\lim_{\Delta x\rightarrow 0} \frac{{u(x+\Delta x)v(x+\Delta x)-u(x)v(x)}}{\Delta x} \\=\lim_{\Delta x\rightarrow 0} \frac{{u(x+\Delta x)v(x+\Delta x)-u(x)v(x) +u(x)v(x+\Delta x)-u(x)v(x+\Delta x)}} {\Delta x} \\配凑\&组合 \\ =\lim_{\Delta x\rightarrow 0} \frac{[(u(x+\Delta x)-u(x))v(x+\Delta x)] +[u(x)(v(x+\Delta x)-v(x))]} {\Delta x} \\=\lim\limits{\Delta x\rightarrow 0} \left \{ {\frac{(u(x+\Delta x)-u(x))}{\Delta x}\cdot v(x+\Delta x) +u(x)\cdot \frac{(v(x+\Delta x)-v(x))}{\Delta x}} \right \} \\=\lim_{\Delta x\rightarrow0} {\frac{(u(x+\Delta x)-u(x))}{\Delta x}\cdot v(x+\Delta x)} +\lim_{\Delta x\rightarrow0}u(x)\cdot \frac{(v(x+\Delta x)-v(x))}{\Delta x} \\=\lim_{\Delta x\rightarrow0} {\frac{(u(x+\Delta x)-u(x))}{\Delta x}\lim_{\Delta x\rightarrow0} v(x+\Delta x)} \\ +\lim_{\Delta x\rightarrow0}u(x)\lim_{\Delta x\rightarrow0} \frac{(v(x+\Delta x)-v(x))}{\Delta x} \\=u'(x)v(x)+u(x)v'(x)
f(x)=u(x)v(x)f′(x)=Δx→0limΔxf(x+Δx)−f(x)=Δx→0limΔxu(x+Δx)v(x+Δx)−u(x)v(x)=Δx→0limΔxu(x+Δx)v(x+Δx)−u(x)v(x)+u(x)v(x+Δx)−u(x)v(x+Δx)配凑&组合=Δx→0limΔx[(u(x+Δx)−u(x))v(x+Δx)]+[u(x)(v(x+Δx)−v(x))]=limΔx→0{Δx(u(x+Δx)−u(x))⋅v(x+Δx)+u(x)⋅Δx(v(x+Δx)−v(x))}=Δx→0limΔx(u(x+Δx)−u(x))⋅v(x+Δx)+Δx→0limu(x)⋅Δx(v(x+Δx)−v(x))=Δx→0limΔx(u(x+Δx)−u(x))Δx→0limv(x+Δx)+Δx→0limu(x)Δx→0limΔx(v(x+Δx)−v(x))=u′(x)v(x)+u(x)v′(x)