MIT_单变量微积分_06

指数与对数

1. 指数

符号:

$a>0时，a^0=1,a^1=a,a^2=a\cdot a.$

基础公式： $a^{x_1+x_2}=a^{x_1}\cdot a^{x_2},{(a^{x_1})}^{x_2}=a^{x_1{x}_2}$.

几何图形：

Ex: 求$\frac{d}{dx}{a}^x等于多少？$
$$\begin{gathered} \frac{d}{dx}{a}^x=\underset{\Delta x\to 0}{\lim }\frac{a^{x+\Delta x}-a^x}{\Delta x} \\ =a^x\underset{\Delta x\to 0}{\lim }\frac{a^{\Delta x}-1}{\Delta x} \end{gathered}$$

令$M(a)={\lim }_{\Delta x\to 0}\frac{a^{\Delta x}-1}{\Delta x},则\frac{d}{dx}{a}^x=M(a)a^x$.

因此，$x=0时，\frac{d}{dx}{a}^x=M(a)$,得出$M(a)$是所有指数函数在$x=0$是的斜率。

假设$M(e)=1$.则$\frac{d}{dx}{e}^x=e^x$,符合上述结论$\frac{d}{dx}{e}^x{|}_{x=0}=1$

• 为什么e存在？

证明：
$$\begin{gathered} f(x)=2^x,f(0{)}^{\prime }=M(2) \\ 扩展k倍后得出， \\ f(kx)=2^{kx}=(2^k{)}^x=b^x,b=2^k \\ \frac{d}{dx}{b}^x=\frac{d}{dx}f(kx)=k{f}^{\prime }(kx) \\ \frac{d}{dx}{b}^x{|}_{x=0}=k{f}^{\prime }(0)=kM(2) \\ b=e\to k=\frac{1}{M(2)} \end{gathered}$$

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Proof 1:
$$\begin{gathered} (e^{lna}{)}^x=a^x,则 \\ \frac{d}{dx}{e}^{xlna}=lna\cdot e^{xlna} \\ =lna\cdot a^x \end{gathered}$$
所以：$M(a)=lna$

Proof 2 对数微分法
令$u=a^x$
$$\begin{gathered} lnu=xlna \\ \frac{u^{\prime }}{u}=l{n}^{\prime }u=lna \\ {u}^{\prime }=ulna \\ \frac{d}{dx}{a}^x=a^{x}lna \end{gathered}$$

2. 自然对数

符号: $w=lnx,y=e^x\iff x=lny$
性质: $ln(x_1\ast x_2)=ln{x}_1+ln{x}_2,ln1=0,lne=1$
几何图形：

Ex:求$lnx$的导数（已知$e^x$的导数求你函数的导数）？
$w=lnx,e^w=x,e^{lnx}=x$
$$\begin{gathered} \frac{d}{dx}{e}^w=\frac{d}{dx}x=1 \\ \frac{d}{dw}{e}^w\frac{d}{dx}w=1 \\ {e}^w\frac{d}{dx}w=1 \\ {w}^{\prime }=\frac{1}{e^w} \\ {w}^{\prime }=\frac{1}{x} \\ l{n}^{\prime }x=\frac{1}{x} \end{gathered}$$

此时，再去求$\frac{d}{dx}{a}^x$的导数？。

2.1 对数微分法

Ex求$\frac{d}{dx}lnu=?$
$$\begin{gathered} \frac{d}{dx}lnu=\frac{d}{du}lnu\frac{d}{dx}u \\ =\frac{1}{u}\cdot \frac{du}{dx} \\ =\frac{u^{\prime }}{u} \end{gathered}$$
所以$\frac{d}{dx}lnu=\frac{u^{\prime }}{u}$

2.2 变动的指数函数求导

Ex:
$v=x^x,lnv=xlnx$
$$\begin{gathered} l{n}^{\prime }v=lnx+x\cdot \frac{1}{x} \\ \frac{v^{\prime }}{v}=1+lnx \\ {v}^{\prime }=v(1+lnx) \\ \frac{d}{dx}{x}^x=x^x(1+lnx) \end{gathered}$$

Ex 求${\lim }_{n\to \infty }(1+\frac{1}{n}{)}^n=?$
提示：$\frac{1}{n}=\Delta x\to 0$
$$\begin{gathered} ln((1+\frac{1}{n}))=nln(1+\frac{1}{n}) \\ =\frac{1}{\Delta x}ln(1+\Delta x) \\ =\frac{1}{\Delta x}ln(1+\Delta x)-ln1 \\ =\frac{d}{dx}lnx{|}_{x=1}=1 \end{gathered}$$

Ex$求证\frac{d}{dx}{x}^r=r{x}^{r-1}$
Proof 1:$x^r=e^{rlnx}$
$$\begin{gathered} \frac{d}{dx}{x}^r=\frac{d}{dx}{e}^{rlnx} \\ =e^{rlnx}\cdot \frac{r}{x} \\ =r{x}^{r-1} \end{gathered}$$
Proof 2 $u=x^r,lnu=rlnx$
$$\begin{gathered} \frac{u^{\prime }}{u}=l{n}^{\prime }u=\frac{r}{x} \\ {u}^{\prime }=u\cdot \frac{r}{x} \\ =r{x}^{r-1} \end{gathered}$$