标签：alignedat frac 浅谈 导数 ln lim Delta sin

本文知识部分由AKauto 和 mashduihca倾情提供

目录

1. 导数表及其证明

2. 导数运算法则及其证明

3. 练习题

前言

函数的导数是表示函数在某一点的切线斜率的函数。

前置知识：

\[\lim_{x\to \infty}e=(1+\frac{1}{x})^x \]

\[\lim_{x\to 0}\frac{\sin x}{x}=1 \]

导数表及其证明

1. \(f(x)=a,f'(x)=0\)(不会有人这个都看不懂吧

2. \(f(x)=x^n,f'(x)=nx^{n-1}\)

\(\begin{alignedat}{6}f'(x)&=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\&=\lim_{\Delta x \to 0}\frac{\sum_{i=0}^n\binom n i x^i\Delta x^{n-i}-x^n}{\Delta x} \\&=\lim_{\Delta x \to 0}\frac{\sum_{i=0}^{n-1}\binom n i x^i\Delta x^{n-i}}{\Delta x} \\&=\lim_{\Delta x \to 0}n\sum_{i=0}^{n-1}\binom {n-1} i x^i\Delta x^{n-i-1} \\&=\lim_{\Delta x \to 0}n(x+\Delta x)^{n-1} \\&=nx^{n-1}\end{alignedat}\)

3. \(f(x)=x^a,f'(x)=ax^{a-1}\) (不会证

4. \(f(x)=e^x,f'(x)=e^x\)

\[\lim_{x\to \infty}e=(1+\frac{1}{x})^x \]

\[\lim_{x\to 0}e=(1+x)^\frac{1}{x} \]

\[\lim_{x\to 0}e^x=1+x \]

\[\lim_{x\to 0}\frac{e^x-1}{x}=1 \]

\(\begin{alignedat}{6}f'(x)&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{e^{x+\Delta x}-e^x}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{e^x\cdot (e^{\Delta x}-1)}{\Delta x} \\&=e^x\end{alignedat}\)

5. \(f(x)=a^x,f'(x)=\ln a\cdot a^x\)

\[\lim_{x\to \infty}e=(1+\frac{1}{x})^x \]

设 \(k=\ln a\)

\[\lim_{x\to 0}e=(1+kx)^\frac{1}{kx} \]

\[\lim_{x\to 0}e^{kx}=1+kx \]

\[\lim_{x\to 0}\frac{a^{x}-1}{x}=\ln a \]

\(\begin{alignedat}{5}f'(x)&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{a^{x+\Delta x}-a^x}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{a^x\cdot (a^{\Delta x}-1)}{\Delta x} \\&=\ln a\cdot a^x\end{alignedat}\)

6. \(f(x)=\ln x,f'(x)=\frac{1}{x}\)

\(\begin{alignedat}{5}f'(x)&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\ln(x+\Delta x)-\ln x}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\ln(1+\frac{\Delta x}{x})}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\frac{\Delta x}{x} \ln (1+\frac{\Delta x}{x})^\frac{x}{\Delta x}}{\Delta x} \\&=\frac{1}{x}\end{alignedat}\)

7. \(f(x)=\log_a x,f'(x)=\frac{1}{x\ln a}\)

\(\begin{alignedat}{5}f'(x)&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\log_a(x+\Delta x)-\log_a x}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\log_a(1+\frac{\Delta x}{x})}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\frac{\Delta x}{x} \ln (1+\frac{\Delta x}{x})^\frac{x}{\Delta x}}{\Delta x\ln a} \\&=\frac{1}{x\ln a}\end{alignedat}\)

8. \(f(x)=\sin x,f'(x)=\cos x\)

\(\begin{alignedat}{5}f'(x)&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\sin(x+\Delta x)-\sin x}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\sin x\cos \Delta x+\cos x\sin\Delta x-\sin x}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\cos x\sin\Delta x}{\Delta x} \\&=\cos x\end{alignedat}\)

9. \(f(x)=\cos x,f'(x)=-\sin x\)

\(\begin{alignedat}{5}f'(x)&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\cos(x+\Delta x)-\cos x}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{cos\ x\cos \Delta x-\sin x\sin\Delta x-\cos x}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{-\sin x\sin\Delta x}{\Delta x} \\&=-\sin x\end{alignedat}\)

导数运算法则及其证明

1. \(h(x)=f(x)+g(x),h'(x)=f'(x)+g'(x)\)

\(\begin{alignedat}{5}h'(x)&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)+g(x+\Delta x)-f(x)-g(x)}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}+\frac{g(x+\Delta x)-g(x)}{\Delta x} \\&=f'(x)+g'(x)\end{alignedat}\)

2. \(h(x)=f(x)-g(x),h'(x)=f'(x)-g'(x)\)

读者自证不难。

3. \(h(x)=f(x)g(x),h'(x)=f'(x)g(x)+g'(x)f(x)\)

\(\begin{alignedat}{5}h'(x)&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)+f(x)(g+\Delta x)-f(x)g(x)}{\Delta x} \\&=f'(x)g(x)+g'(x)f(x)\end{alignedat}\)

4. \(h(x)=\frac{f(x)}{g(x)},h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\)

\(\begin{alignedat}{5}h'(x)&=\lim_{\Delta x\to 0}\frac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x)-f(x)g(x+\Delta x)}{g(x+\Delta x)g(x)\Delta x} \\&=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+\Delta x)}{g(x+\Delta x)g(x)\Delta x} \\&=\lim_{\Delta x\to 0}\frac{f'(x)g(x)-f(x)g'(x)}{g(x+\Delta x)g(x)} \\&=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\end{alignedat}\)

5. \(h(x)=f(g(x)),h'(x)=f'(g(x))g'(x)\)

\(\begin{alignedat}{5}h'(x)&=\lim_{\Delta x\to 0}\frac{f(g(x+\Delta x))-f(g(x))}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{f(g(x+\Delta x))-f(g(x))}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{f(g(x)+\Delta x\cdot g'(x))-f(g(x))}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{f(g(x))+\Delta x\cdot g'(x)f'(g(x))-f(g(x))}{\Delta x} \\&=\lim_{\Delta x\to 0}\frac{\Delta x\cdot g'(x)f'(g(x))}{\Delta x} \\&=f'(g(x))g'(x)\end{alignedat}\)

练习题

1. 求 \(f(x)=x^{x^x}\) 的导数

则\(f(x)=x^{x^x}=x^{(x^2)}\)

设 \(g(x)=\ln f(x)=\ln x^{(x^2)}=x^2\ln x\)

\(g'(x)=(x^2\ln x)'=2x\ln x+x\)

\(g'(x)=(\ln f(x))'=\ln'f(x)\cdot f'(x)=\frac{f'(x)}{f(x)}\)

所以 \(\frac{f'(x)}{f(x)}=2x\ln x+x\)

\(f'(x)=2x\ln x\cdot x^{x^x}+x\cdot x^{x^x}\)

标签：alignedat,frac,浅谈,导数,ln,lim,Delta,sin
From： https://www.cnblogs.com/mikefeng/p/17367957.html