数论基础——求导
一、定义
对于函数 \(f(x)=1,x,x+1,2^x,...\),我们定义 \(f'(x)=\displaystyle\lim_{\Delta_x\rightarrow0}\frac{f'(x+\Delta_x)-f(x)}{\Delta_x}\)。
二、常见函数求导
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\(f(x)=1\)
\[\begin{align*} f'(x) &=\lim_{\Delta_x\rightarrow0}\frac{f(x+\Delta_x)-f(x)}{\Delta_x}\\ &=\lim_{\Delta_x\rightarrow0}\frac{1-1}{\Delta_x}\\ &=0 \end{align*} \] -
\(f(x)=x\)
\[\begin{align*} f'(x) &=\lim_{\Delta_x\rightarrow0}\frac{f(x+\Delta_x)-f(x)}{\Delta_x}\\ &=\lim_{\Delta_x\rightarrow0}\frac{x+\Delta_x-x}{\Delta_x}\\ &=1 \end{align*} \] -
\(f(x)=x^2\)
\[\begin{align*} f'(x) &=\lim_{\Delta_x\rightarrow0}\frac{f(x+\Delta_x)-f(x)}{\Delta_x}\\ &=\lim_{\Delta_x\rightarrow0}\frac{(x+\Delta_x)^2-x^2}{\Delta_x}\\ &=\lim_{\Delta_x\rightarrow0}\frac{x^2+2x\Delta_x+\Delta_x^2-x^2}{\Delta_x}\\ &=2x \end{align*} \] -
\(f(x)=x^n\)
\[\begin{align*} f'(x) &=\lim_{\Delta_x\rightarrow0}\frac{f(x+\Delta_x)-f(x)}{\Delta_x}\\ &=\lim_{\Delta_x\rightarrow0}\frac{(x+\Delta_x)^n-x^n}{\Delta_x}\\ &=\lim_{\Delta_x\rightarrow0}\frac{\displaystyle\sum_{k=0}^n[{n\choose k}\Delta_x^kx^{n-k}]-x^n}{\Delta_x}\\ &=\lim_{\Delta_x\rightarrow0}\frac{\displaystyle\sum_{k=1}^n[{n\choose k}\Delta_x^kx^{n-k}]}{\Delta_x}\\ &=\lim_{\Delta_x\rightarrow0}{\displaystyle\sum_{k=1}^n[{n\choose k}\Delta_x^{k-1}x^{n-k}]}\\ &={n\choose 1}x^{n-1}\\ &=nx^{n-1} \end{align*} \] -
\(f(x)=\sin x\)
\[\begin{align*} & \sin(x+\Delta)-\sin x\\ =&\sin x\cos \Delta+\cos x\sin \Delta-\sin x\\ =&\sin x\cdot1+\cos x\sin \Delta -\sin x \end{align*} \]则:
\[\begin{align*} f'(x)&=\frac{\cos x\sin \Delta}{\Delta}\\ &=\cos x \end{align*} \]
三、法则
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\(h(x)=f(x)+g(x)\)
\[\begin{align*} h'(x)&=\frac{f(x+\Delta)+g(x+\Delta)-f(x)-g(x)}{\Delta}\\ &=f'(x)+g'(x) \end{align*} \] -
\(h(x)=c\cdot f(x)\)
\[\begin{align*} h'(x)&=\frac{c\cdot f(x+\Delta)-c\cdot f(x)}{\Delta}\\ &=c\cdot f'(x) \end{align*} \] -
\(h(x)=f(x)\cdot g(x)\)
\[\begin{align*} h'(x)&=\frac{f(x+\Delta)\cdot g(x+\Delta)-f(x)\cdot g(x)}{\Delta}\\ &=\frac{f(x+\Delta)\cdot g(x+\Delta)-f(x)\cdot g(x)-f(x+\Delta)g(x)+f(x+\Delta)g(x)}{\Delta}\\ &=\frac{f(x+\Delta)\cdot(g(x+\Delta)-g(x))+g(x)\cdot (f(x+\Delta)-f(x))}{\Delta}\\ &=f'(x)g(x)+f(x)g'(x)\\ \end{align*} \] -
\(h(x)=f(g(x))\)
\[\begin{align*} h'(x)&=\frac{f(g(x+\Delta))-f(g(x))}{\Delta}\\ &=\frac{(f(g(x+\Delta))-f(g(x)))\times(g(x+\Delta)-g(x))}{\Delta\times(g(x+\Delta)-g(x))}\\ &=\frac{f(g(x+\Delta))-f(g(x))}{g'(x)}\cdot{g'(x)}\\ &=f'(g(x))g'(x) \end{align*} \]

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