完整推導(dǎo)導(dǎo)數(shù)公式體系（1）

2022-02-05 09:08 作者:匆匆-cc 0人讀過(guò) | 我要投稿

模塊零：一些工具

????????工具一：導(dǎo)數(shù)的定義。

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

????????工具二：兩個(gè)重要極限。

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

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

????????第一個(gè)極限證明：

????????????高中推導(dǎo)過(guò)，在 $(0%2C%5Cfrac%7B%5Cpi%7D%7B2%7D)$ 上有

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

????????????事實(shí)上，我們有

????????????推導(dǎo)1：考察三角函數(shù)線，在單位圓中，我們會(huì)發(fā)現(xiàn)，正弦線 $%5Cvec%7BDA%7D$ 的大小隨著角度逐漸減小會(huì)逐漸接近于弧長(zhǎng)，這表明： $%5Csin%20x%20%5Csim%20x$ 。

????????????推導(dǎo)2：考察圖形面積，

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

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

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

????????????注意到

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

????????????所以

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

????????????考慮到當(dāng)角度越來(lái)越小時(shí)，扇形的面積越來(lái)越接近于三角形AOC的面積。所以有 $%5Csin%20x%20%5Csim%20x$ 。

????## 圖為 $%5Ccolor%7Bgray%7D%7B%5Cfrac%7B%5Csin%20x%7D%7Bx%7D%7D$ 圖像

????????第二個(gè)極限證明：

????????????關(guān)于 $%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5En$ 的單調(diào)性可參考下面鏈接。

????????????鏈接中直接給出該式的極限是 $e$ ，下面補(bǔ)充 $n%5Cto%20%2B%5Cinfty$ 對(duì)其上界的證明。

????????????根據(jù)牛頓二項(xiàng)式定理，當(dāng) $n$ 為正整數(shù)時(shí)，我們有

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

????????????所以該式有上界。

????????????根據(jù)單調(diào)有界數(shù)列必有極限得到，該式有極限。

????????????我們用字母 $e$ 來(lái)表示該極限。（當(dāng)然其實(shí)到這里 $e$ 還是沒有算出來(lái)，但是仍然可以通過(guò)泰勒級(jí)數(shù)展開來(lái)計(jì)算，因?yàn)檫@時(shí)我們已經(jīng)承認(rèn)了 $%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cleft(1%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5En%20%3De$ ）

????????????## 其實(shí)這一切都可以通過(guò) $%5Ccolor%7Bgray%7De$ 的泰勒展開來(lái)說(shuō)明，但是從嚴(yán)謹(jǐn)性的角度來(lái)說(shuō)，此時(shí)不應(yīng)該出現(xiàn)泰勒展開，因此這里采用了 $%5Ccolor%7Bgray%7D%7Bn%7D$ 為正整數(shù)時(shí)的牛頓二項(xiàng)式定理展開。

????????????另外，還有一個(gè)常用的等價(jià)無(wú)窮小。

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

????????工具三：函數(shù)求導(dǎo)法則

????????????①函數(shù)的和差積商求導(dǎo)法則

????????????和、差：

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

????????????積：

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

????????????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反函數(shù)的求導(dǎo)法則

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

????????????也就是說(shuō)，反函數(shù)的導(dǎo)數(shù)等于原函數(shù)的導(dǎo)數(shù)的倒數(shù)。

????????????③復(fù)合函數(shù)的求導(dǎo)法則

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

????????????這只是簡(jiǎn)單證明，存在不嚴(yán)格之處。（分子分母同乘的數(shù)可能為0）

????????????④隱函數(shù)的求導(dǎo)法則（微分）

????????????簡(jiǎn)單來(lái)說(shuō)，就是等式兩邊同時(shí)對(duì) $x$ 求導(dǎo)。

模塊一：常函數(shù)

$f(x)%3DC$

模塊二：冪函數(shù)

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

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

????????以上證明限于 $%5Calpha$ 為正整數(shù)的情形。

????????下面證明 $%5Calpha$ 為實(shí)數(shù)的情形。

????## 這里用到了等價(jià)無(wú)窮小

模塊三：對(duì)數(shù)函數(shù)

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

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