完整推导导数公式体系（1）

2022-02-05 09:08 作者:匆匆-cc 0人读过 | 我要投稿

模块零：一些工具

工具一：导数的定义。

$f'(x)%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%20%5Cfrac%7Bf(x%2B%5CDelta%20x)-f(x)%7D%7B%5CDelta%20x%7D%20$

工具二：两个重要极限。

$%5Clim_%7Bx%5Cto0%7D%20%5Cfrac%7B%5Csin%20x%7D%7Bx%7D%20%3D1$

$%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5En%20%3De$

第一个极限证明：

高中推导过，在 $(0%2C%5Cfrac%7B%5Cpi%7D%7B2%7D)$ 上有

$%5Csin%20x%3Cx%3C%5Ctan%20x$

事实上，我们有

推导1：考察三角函数线，在单位圆中，我们会发现，正弦线 $%5Cvec%7BDA%7D$ 的大小随着角度逐渐减小会逐渐接近于弧长，这表明： $%5Csin%20x%20%5Csim%20x$ 。

推导2：考察图形面积，

$S_%7B%5Ctriangle%20AOC%7D%3D%5Cfrac%7B1%7D%7B2%7DAD%5Ccdot%20OC%3D%5Cfrac%7B1%7D%7B2%7D%5Csin%20x%5Ccdot%201%3D%5Cfrac%7B1%7D%7B2%7D%5Csin%20x$

$S_%7B%E6%89%87AOC%7D%3D%5Cfrac%7B1%7D%7B2%7Dx%5Ccdot%201%5E2%3D%5Cfrac%7B1%7D%7B2%7Dx$

$S_%7B%5Ctriangle%20BOC%7D%3D%5Cfrac%7B1%7D%7B2%7DBC%5Ccdot%20OC%3D%5Cfrac%7B1%7D%7B2%7D%5Ctan%20x%5Ccdot%201%3D%5Cfrac%7B1%7D%7B2%7D%5Ctan%20x$

注意到

$S_%7B%5Ctriangle%20AOC%7D%3CS_%7B%E6%89%87AOC%7D%3CS_%7B%5Ctriangle%20BOC%7D$

所以

$%5Csin%20x%3Cx%3C%5Ctan%20x$

考虑到当角度越来越小时，扇形的面积越来越接近于三角形AOC的面积。所以有 $%5Csin%20x%20%5Csim%20x$ 。

## 图为 $%5Ccolor%7Bgray%7D%7B%5Cfrac%7B%5Csin%20x%7D%7Bx%7D%7D$ 图像

第二个极限证明：

关于 $%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5En$ 的单调性可参考下面链接。

链接中直接给出该式的极限是 $e$ ，下面补充 $n%5Cto%20%2B%5Cinfty$ 对其上界的证明。

根据牛顿二项式定理，当 $n$ 为正整数时，我们有

$%5Cbegin%7Balign%7D%0A%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5En%0A%26%3D1%2B%5Cfrac%7Bn%7D%7B1!%7D%5Ccdot%5Cfrac%7B1%7D%7Bn%7D%2B%5Cfrac%7Bn(n-1)%7D%7B2!%7D%5Ccdot%5Cfrac%7B1%7D%7Bn%5E2%7D%2B%5Cfrac%7Bn(n-1)(n-2)%7D%7B3!%7D%5Ccdot%5Cfrac%7B1%7D%7Bn%5E3%7D%2B%E2%80%A6%2B%5Cfrac%7Bn(n-1)%E2%80%A6(n-n%2B1)%7D%7Bn!%7D%5Ccdot%5Cfrac%7B1%7D%7Bn%5En%7D%0A%5C%5C%26%3D1%2B1%2B%5Cfrac%7B1%7D%7B2!%7D%5Cleft(1-%5Cfrac%7B1%7D%7Bn%7D%5Cright)%2B%5Cfrac%7B1%7D%7B3!%7D%5Cleft(1-%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5Cleft(1-%5Cfrac%7B2%7D%7Bn%7D%5Cright)%2B%E2%80%A6%2B%5Cfrac%7B1%7D%7Bn!%7D%5Cleft(1-%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5Cleft(1-%5Cfrac%7B2%7D%7Bn%7D%5Cright)%E2%80%A6%5Cleft(1-%5Cfrac%7Bn-1%7D%7Bn%7D%5Cright)%0A%5C%5C%26%3C1%2B1%2B%5Cfrac%7B1%7D%7B2!%7D%2B%5Cfrac%7B1%7D%7B3!%7D%2B%E2%80%A6%2B%5Cfrac%7B1%7D%7Bn!%7D%0A%5C%5C%26%5Cleq%201%2B%5Cleft(1%2B%5Cfrac%7B1%7D%7B2%7D%2B%5Cfrac%7B1%7D%7B2%5E2%7D%2B%E2%80%A6%2B%5Cfrac%7B1%7D%7B2%5E%7Bn-1%7D%7D%5Cright)%0A%5C%5C%26%3D1%2B%5Cfrac%7B1-%5Cfrac%7B1%7D%7B2%5En%7D%7D%7B1-%5Cfrac%7B1%7D%7B2%7D%7D%0A%5C%5C%26%3D3-%5Cfrac%7B1%7D%7B2%5E%7Bn-1%7D%7D%0A%5C%5C%26%3C3%0A%5Cend%7Balign%7D$

所以该式有上界。

根据单调有界数列必有极限得到，该式有极限。

我们用字母 $e$ 来表示该极限。（当然其实到这里 $e$ 还是没有算出来，但是仍然可以通过泰勒级数展开来计算，因为这时我们已经承认了 $%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5En%20%3De$ ）

## 其实这一切都可以通过 $%5Ccolor%7Bgray%7De$ 的泰勒展开来说明，但是从严谨性的角度来说，此时不应该出现泰勒展开，因此这里采用了 $%5Ccolor%7Bgray%7D%7Bn%7D$ 为正整数时的牛顿二项式定理展开。

另外，还有一个常用的等价无穷小。

$%5Cbegin%7Balign%7D%0A%5Clim_%7Bx%5Cto0%7D%5Cfrac%7B%5Csqrt%5Bn%5D%7B1%2Bx%7D-1%7D%7B%5Cfrac%7B1%7D%7Bn%7Dx%7D%26%3D%5Clim_%7Bx%5Cto0%7D%5Cfrac%7B(%5Csqrt%5Bn%5D%7B1%2Bx%7D)%5En-1%7D%7B%5Cfrac%7B1%7D%7Bn%7Dx%5B%5Csqrt%5Bn%5D%7B(1%2Bx)%5E%7Bn-1%7D%7D%2B%5Csqrt%5Bn%5D%7B(1%2Bx)%5E%7Bn-2%7D%7D%2B%E2%80%A6%2B1%5D%7D%0A%5C%5C%26%3D%5Clim_%7Bx%5Cto0%7D%5Cfrac%7Bn%7D%7B%5Csqrt%5Bn%5D%7B(1%2Bx)%5E%7Bn-1%7D%7D%2B%5Csqrt%5Bn%5D%7B(1%2Bx)%5E%7Bn-2%7D%7D%2B%E2%80%A6%2B1%7D%0A%5C%5C%26%3D%5Cfrac%7Bn%7D%7Bn%7D%0A%5C%5C%26%3D1%0A%5Cend%7Balign%7D$

工具三：函数求导法则

①函数的和差积商求导法则

和、差：

$%5Cbegin%7Balign%7D%0A%5Bu(x)%5Cpm%20v(x)%5D'%26%3D%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cfrac%7B%5Bu(x%2B%5CDelta%20x)%5Cpm%20v(x%2B%5CDelta%20x)%5D-%5Bu(x)%5Cpm%20v(x)%5D%7D%7B%5CDelta%20x%7D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cfrac%7Bu(x%2B%5CDelta%20x)-u(x)%7D%7B%5CDelta%20x%7D%5Cpm%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cfrac%7Bv(x%2B%5CDelta%20x)-v(x)%7D%7B%5CDelta%20x%7D%0A%5C%5C%26%3Du'(x)%5Cpm%20v'(x)%0A%5Cend%7Balign%7D$

积：

$%5Cbegin%7Balign%7D%0A%5Bu(x)v(x)%5D'%26%3D%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cfrac%7Bu(x%2B%5CDelta%20x)v(x%2B%5CDelta%20x)-u(x)v(x)%7D%7B%5CDelta%20x%7D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cleft%5B%5Cfrac%7Bu(x%2B%5CDelta%20x)-u(x)%7D%7B%5CDelta%20x%7D%5Ccdot%20v(x%2B%5CDelta%20x)%2Bu(x)%5Ccdot%5Cfrac%7Bv(x%2B%5CDelta%20x)-v(x)%7D%7B%5CDelta%20x%7D%5Cright%5D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cfrac%7Bu(x%2B%5CDelta%20x)-u(x)%7D%7B%5CDelta%20x%7D%5Ccdot%20%5Clim_%7B%5CDelta%20x%5Cto0%7Dv(x%2B%5CDelta%20x)%2Bu(x)%5Ccdot%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cfrac%7Bv(x%2B%5CDelta%20x)-v(x)%7D%7B%5CDelta%20x%7D%0A%5C%5C%26%3Du'(x)v(x)%2Bu(x)v'(x)%0A%5Cend%7Balign%7D$

商：

$%5Cbegin%7Balign%7D%0A%5Cleft%5B%5Cfrac%7Bu(x)%7D%7Bv(x)%7D%5Cright%5D'%26%3D%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cfrac%7B%5Cfrac%7Bu(x%2B%5CDelta%20x)%7D%7Bv(x%2B%5CDelta%20x)%7D-%5Cfrac%7Bu(x)%7D%7Bv(x)%7D%7D%7B%5CDelta%20x%7D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cfrac%7Bu(x%2B%5CDelta%20x)v(x)-u(x)v(x%2B%5CDelta%20x)%7D%7Bv(x%2B%5CDelta%20x)v(x)%5CDelta%20x%7D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cfrac%7B%5Bu(x%2B%5CDelta%20x)-u(x)%5Dv(x)-u(x)%5Bv(x%2B%5CDelta%20x)-v(x)%5D%7D%7Bv(x%2B%5CDelta%20x)v(x)%5CDelta%20x%7D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%5Cto0%7D%5Cfrac%7B%5Cfrac%7Bu(x%2B%5CDelta%20x)-u(x)%7D%7B%5CDelta%20x%7Dv(x)-u(x)%5Cfrac%7Bv(x%2B%5CDelta%20x)-v(x)%7D%7B%5CDelta%20x%7D%7D%7Bv(x%2B%5CDelta%20x)v(x)%7D%0A%5C%5C%26%3D%5Cfrac%7Bu'(x)v(x)-u(x)v'(x)%7D%7Bv%5E2(x)%7D%0A%5Cend%7Balign%7D$

②反函数的求导法则

$%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B1%7D%7B%5Cfrac%7Bdx%7D%7Bdy%7D%7D$

也就是说，反函数的导数等于原函数的导数的倒数。

③复合函数的求导法则

$%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7Bdy%7D%7Bdu%7D%5Ccdot%5Cfrac%7Bdu%7D%7Bdx%7D$

这只是简单证明，存在不严格之处。（分子分母同乘的数可能为0）

④隐函数的求导法则（微分）

简单来说，就是等式两边同时对 $x$ 求导。

模块一：常函数

$f(x)%3DC$

模块二：幂函数

$f(x)%3Dx%5E%5Calpha$

$%5Cbegin%7Balign%7D%0Af'(x)%26%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%20%5Cfrac%7Bf(x%2B%5CDelta%20x)-f(x)%7D%7B%5CDelta%20x%7D%20%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%20%5Cfrac%7B(x%2B%5CDelta%20x)%5E%5Calpha-x%5E%5Calpha%7D%7B%5CDelta%20x%7D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%20%5Cfrac%7Bx%5E%5Calpha%2B%5Calpha%20x%5E%7B%5Calpha-1%7D%5CDelta%20x%2B%5Cfrac%7B%5Calpha(%5Calpha-1)%7D%7B2%7Dx%5E%7B%5Calpha-2%7D(%5CDelta%20x)%5E2%2B%E2%80%A6%2B(%5CDelta%20x)%5E%5Calpha-x%5E%5Calpha%7D%7B%5CDelta%20x%7D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%20%5Cfrac%7B%5Calpha%20x%5E%7B%5Calpha-1%7D%5CDelta%20x%2B%5Cfrac%7B%5Calpha(%5Calpha-1)%7D%7B2%7Dx%5E%7B%5Calpha-2%7D(%5CDelta%20x)%5E2%2B%E2%80%A6%2B(%5CDelta%20x)%5E%5Calpha%7D%7B%5CDelta%20x%7D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%20%5Cleft%5B%5Calpha%20x%5E%7B%5Calpha-1%7D%2B%5Cfrac%7B%5Calpha(%5Calpha-1)%7D%7B2%7Dx%5E%7B%5Calpha-2%7D%5CDelta%20x%2B%E2%80%A6%2B(%5CDelta%20x)%5E%7B%5Calpha-1%7D%5Cright%5D%0A%5C%5C%26%3D%5Calpha%20x%5E%7B%5Calpha-1%7D%0A%5Cend%7Balign%7D$

以上证明限于 $%5Calpha$ 为正整数的情形。

下面证明 $%5Calpha$ 为实数的情形。

## 这里用到了等价无穷小

模块三：对数函数

$f(x)%3D%5Clog_a%20x%20$

$%5Cbegin%7Balign%7D%0Af'(x)%26%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%20%5Cfrac%7Bf(x%2B%5CDelta%20x)-f(x)%7D%7B%5CDelta%20x%7D%20%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%20%5Cfrac%7B%5Clog_a(x%2B%5CDelta%20x)-%5Clog_ax%7D%7B%5CDelta%20x%7D%20%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%5Cfrac%7B%5Clog_a%5Cleft(1%2B%5Cfrac%7B%5CDelta%20x%7D%7Bx%7D%5Cright)%7D%7B%5CDelta%20x%7D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%5Clog_a%5Cleft(1%2B%5Cfrac%7B%5CDelta%20x%7D%7Bx%7D%5Cright)%5E%7B%5Cfrac%7B1%7D%7B%5CDelta%20x%7D%7D%0A%5C%5C%26%3D%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%5Clog_a%5Cleft(%5Cleft(1%2B%5Cfrac%7B%5CDelta%20x%7D%7Bx%7D%5Cright)%5E%7B%5Cfrac%7Bx%7D%7B%5CDelta%20x%7D%7D%5Cright)%5E%7B%5Cfrac%7B1%7D%7Bx%7D%7D%0A%5C%5C%26%3D%5Clog_a%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%5Cleft(%5Cleft(1%2B%5Cfrac%7B%5CDelta%20x%7D%7Bx%7D%5Cright)%5E%7B%5Cfrac%7Bx%7D%7B%5CDelta%20x%7D%7D%5Cright)%5E%7B%5Cfrac%7B1%7D%7Bx%7D%7D%0A%5C%5C%26%3D%5Cfrac%7B1%7D%7Bx%7D%5Clog_a%5Clim_%7B%5CDelta%20x%20%5Cto%200%7D%5Cleft(1%2B%5Cfrac%7B%5CDelta%20x%7D%7Bx%7D%5Cright)%5E%7B%5Cfrac%7Bx%7D%7B%5CDelta%20x%7D%7D%0A%5C%5C%26%3D%5Cfrac%7B1%7D%7Bx%7D%5Clog_ae%5C%23%0A%5C%5C%26%3D%5Cfrac%7B1%7D%7Bx%5Cln%20a%7D%0A%5Cend%7Balign%7D$