根据导数的定义 d d x a x = lim ⁡ Δ x → 0 a x + Δ x − a x Δ x = lim ⁡ Δ x → 0 a x ⋅ a Δ x − a x Δ x = a x lim ⁡ Δ x → 0 a Δ x − 1 Δ x \frac{d}{dx}a^x=\underset{\Delta x\rightarrow 0}{\lim}\frac{a^{x+\Delta x}-a^x}{\Delta x}=\underset{\Delta x\rightarrow 0}{\lim}\frac{a^x\cdot a^{\Delta x}-a^x}{\Delta x}=a^x\underset{\Delta x\rightarrow 0}{\lim}\frac{a^{\Delta x}-1}{\Delta x} dxd​ax=Δx→0lim​Δxax+Δx−ax​=Δx→0lim​Δxax⋅aΔx−ax​=axΔx→0lim​ΔxaΔx−1​ 令 M ( a ) = lim ⁡ Δ x → 0 a Δ x − 1 Δ x \text{令}M\left( a \right) =\underset{\Delta x\rightarrow 0}{\lim}\frac{a^{\Delta x}-1}{\Delta x} 令M(a)=Δx→0lim​ΔxaΔx−1​ 则 d d x a x = a x M ( a ) \text{则}\frac{d}{dx}a^x=a^xM\left( a \right) 则dxd​ax=axM(a) 由于 M ( a ) = lim ⁡ Δ x → 0 a Δ x − 1 Δ x = lim ⁡ Δ x → 0 a 0 + Δ x − a 0 Δ x = d d x a x ∣ x = 0 M\left( a \right) =\underset{\Delta x\rightarrow 0}{\lim}\frac{a^{\Delta x}-1}{\Delta x}=\underset{\Delta x\rightarrow 0}{\lim}\frac{a^{0+\Delta x}-a^0}{\Delta x}=\left. \frac{d}{dx}a^x \right|_{x=0} M(a)=Δx→0lim​ΔxaΔx−1​=Δx→0lim​Δxa0+Δx−a0​=dxd​ax∣∣∣∣​x=0​ 得 M ( a ) M(a) M(a)为函数图像 y = a x y=a^x y=ax在 x = 0 x=0 x=0处切线的斜率，如图所示 考虑是否存在 b b b使 M ( b ) = 1 M(b)=1 M(b)=1 令 a = 2 a=2 a=2，函数图像如图所示，割线（secant line）经过点 ( 0 , 1 ) (0,1) (0,1)和 ( 1 , 2 ) (1,2) (1,2)，其斜率为1，所以可得 M ( 2 ) < 1 M(2)1 M(4)>1 所以存在 b ∈ ( 2 , 4 ) b\in(2,4) b∈(2,4)使 M ( b ) = 1 M(b)=1 M(b)=1 实际上 M ( e ) = 1 ( e 为 自 然 常 数 ) M(e)=1 (e为自然常数) M(e)=1(e为自然常数) 详见： 谈谈高等数学中自然常数e的来历 The Enigmatic Number e: A History in Verse and Its Uses in the Mathematics Classroom

现在我们有 M ( e ) = 1 M(e)=1 M(e)=1，即可得 M ( e ) = lim ⁡ Δ x → 0 e Δ x − 1 Δ x = 1 ( 实 际 上 此 极 限 也 可 从 e 的 定 义 推 导 可 得 ， 从 而 通 过 导 数 的 定 义 计 算 e x 的 导 数 ) M\left( e \right) =\underset{\Delta x\rightarrow 0}{\lim}\frac{e^{\Delta x}-1}{\Delta x}=1 (实际上此极限也可从e的定义推导可得，从而通过导数的定义计算e^x的导数) M(e)=Δx→0lim​ΔxeΔx−1​=1(实际上此极限也可从e的定义推导可得，从而通过导数的定义计算ex的导数) 可计算出导数 d d x e x = e x \frac{d}{dx}e^x=e^x dxd​ex=ex 回到计算 a x a^x ax，通过将 a x a^x ax化为以 e e e为底的指数，来计算其导数 d d x a x = d d x e x ln ⁡ a = ( ln ⁡ a ) e x ln ⁡ a = a x ln ⁡ a \frac{d}{dx}a^x=\frac{d}{dx}e^{x\ln a}=\left( \ln a \right) e^{x\ln a}=a^x\ln a dxd​ax=dxd​exlna=(lna)exlna=axlna

令    y = ln ⁡ x   则    e y = x \text{令\,\,}y=\ln x\ \text{则\,\,}e^y=x 令y=lnx 则ey=x d d x ( e y = x )   ⇒ e y ⋅ d y d x = 1 \frac{d}{dx}\left( e^y=x \right) \ \Rightarrow e^y\cdot \frac{dy}{dx}=1 dxd​(ey=x) ⇒ey⋅dxdy​=1 d y d x = 1 e y = 1 x \frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x} dxdy​=ey1​=x1​

令    y = log ⁡ a x   则    a y = x \text{令\,\,}y=\log _ax\ \text{则\,\,}a^y=x 令y=loga​x 则ay=x d d x ( a y = x ) ⇒ a y ⋅ ln ⁡ a ⋅ d y d x = 1 \frac{d}{dx}\left( a^y=x \right) \Rightarrow a^y\cdot \ln a\cdot \frac{dy}{dx}=1 dxd​(ay=x)⇒ay⋅lna⋅dxdy​=1 d y d x = 1 a y ⋅ ln ⁡ a = 1 x ln ⁡ a \frac{dy}{dx}=\frac{1}{a^y\cdot \ln a}=\frac{1}{x\ln a} dxdy​=ay⋅lna1​=xlna1​

x x = ( e ln ⁡ x ) x = e x ln ⁡ x x^x=\left( e^{\ln x} \right) ^x=e^{x\ln x} xx=(elnx)x=exlnx d d x x x = d d x e x ln ⁡ x = e x ln ⁡ x ( ln ⁡ x + x ⋅ 1 x ) = x x ( ln ⁡ x + 1 ) \frac{d}{dx}x^x=\frac{d}{dx}e^{x\ln x}=e^{x\ln x}\left( \ln x+x\cdot \frac{1}{x} \right) =x^x\left( \ln x+1 \right) dxd​xx=dxd​exlnx=exlnx(lnx+x⋅x1​)=xx(lnx+1)

x r = ( e ln ⁡ x ) r = e r ln ⁡ x x^r=\left( e^{\ln x} \right) ^r=e^{r\ln x} xr=(elnx)r=erlnx d d x x r = d d x e r ln ⁡ x = e r ln ⁡ x ⋅ r x = x r ⋅ r x = r ⋅ x r − 1 \frac{d}{dx}x^r=\frac{d}{dx}e^{r\ln x}=e^{r\ln x}\cdot \frac{r}{x}=x^r\cdot \frac{r}{x}=r\cdot x^{r-1} dxd​xr=dxd​erlnx=erlnx⋅xr​=xr⋅xr​=r⋅xr−1

d d x sin ⁡ x = lim ⁡ Δ x → 0 sin ⁡ ( x + Δ x ) − sin ⁡ x Δ x = lim ⁡ Δ x → 0 sin ⁡ x ⋅ cos ⁡ Δ x + cos ⁡ x ⋅ sin ⁡ Δ x − sin ⁡ x Δ x   ( 两 角 和 公 式 sin ⁡ ( a + b ) = sin ⁡ a ⋅ cos ⁡ b + cos ⁡ a ⋅ sin ⁡ b ) = lim ⁡ Δ x → 0 sin ⁡ x ⋅ ( cos ⁡ Δ x − 1 ) Δ x + lim ⁡ Δ x → 0 cos ⁡ x ⋅ sin ⁡ Δ x Δ x \begin{aligned} \frac{d}{dx}\sin x&=\underset{\Delta x\rightarrow 0}{\lim}\frac{\sin \left( x+\Delta x \right) -\sin x}{\Delta x} \\ &=\underset{\Delta x\rightarrow 0}{\lim}\frac{\sin x\cdot \cos \Delta x+\cos x\cdot \sin \Delta x-\sin x}{\Delta x}\ \left(两角和公式 \sin \left( a+b \right) =\sin a\cdot \cos b+\cos a\cdot \sin b \right) \\ &=\underset{\Delta x\rightarrow 0}{\lim}\frac{\sin x\cdot \left( \cos \Delta x-1 \right)}{\Delta x}+ \underset{\Delta x\rightarrow 0}{\lim}\frac{\cos x\cdot \sin \Delta x}{\Delta x} \end{aligned} dxd​sinx​=Δx→0lim​Δxsin(x+Δx)−sinx​=Δx→0lim​Δxsinx⋅cosΔx+cosx⋅sinΔx−sinx​ (两角和公式sin(a+b)=sina⋅cosb+cosa⋅sinb)=Δx→0lim​Δxsinx⋅(cosΔx−1)​+Δx→0lim​Δxcosx⋅sinΔx​​ 由于 lim ⁡ x → 0 cos ⁡ x − 1 x = lim ⁡ x → 0 1 − 2 sin ⁡ 2 x 2 − 1 x   ( 倍 角 公 式 cos ⁡ 2 x = 1 − 2 sin ⁡ 2 x ) = − lim ⁡ x → 0 ( sin ⁡ x 2 x 2 ⋅ sin ⁡ x 2 ) = − lim ⁡ x → 0 sin ⁡ x 2   ( 重要极限 lim ⁡ x → 0 sin ⁡ x x = 1 ) = 0 \begin{aligned} \underset{x\rightarrow 0}{\lim}\frac{\cos x-1}{x}&=\underset{x\rightarrow 0}{\lim}\frac{1-2\sin ^2\frac{x}{2}-1}{x}\ \left(倍角公式 \cos 2x=1-2\sin ^2x \right)\\ &=-\underset{x\rightarrow 0}{\lim}\left( \frac{\sin \frac{x}{2}}{\frac{x}{2}}\cdot \sin \frac{x}{2} \right) \\ &=-\underset{x\rightarrow 0}{\lim}\sin \frac{x}{2}\ \left( \text{重要极限}\underset{x\rightarrow 0}{\lim}\frac{\sin x}{x}=1 \right) \\ &=0 \end{aligned} x→0lim​xcosx−1​​=x→0lim​x1−2sin22x​−1​ (倍角公式cos2x=1−2sin2x)=−x→0lim​(2x​sin2x​​⋅sin2x​)=−x→0lim​sin2x​ (重要极限x→0lim​xsinx​=1)=0​ 得 d d x sin ⁡ x = lim ⁡ Δ x → 0 sin ⁡ x ⋅ ( cos ⁡ Δ x − 1 ) Δ x + lim ⁡ Δ x → 0 cos ⁡ x ⋅ sin ⁡ Δ x Δ x = cos ⁡ x \frac{d}{dx}\sin x=\underset{\Delta x\rightarrow 0}{\lim}\frac{\sin x\cdot \left( \cos \Delta x-1 \right)}{\Delta x}+ \underset{\Delta x\rightarrow 0}{\lim}\frac{\cos x\cdot \sin \Delta x}{\Delta x}=\cos x dxd​sinx=Δx→0lim​Δxsinx⋅(cosΔx−1)​+Δx→0lim​Δxcosx⋅sinΔx​=cosx

由于 Δ x \Delta x Δx很小，所以可以认为 O A ⊥ A B OA \bot AB OA⊥AB和 A B = Δ x ( Δ x 为 弧 度 ) AB=\Delta x (\Delta x 为弧度) AB=Δx(Δx为弧度)，易证 ∠ A B E = x \angle ABE=x ∠ABE=x（OA逆时针旋转90°和AB平行，OD逆时针旋转90°和BE平行，OA与OD的夹角为 x x x，则AB与BE的夹角也为 x x x） 可 得 Δ y Δ x = B E A B = cos ⁡ x ， 即 d d x sin ⁡ x = cos ⁡ x 可得\frac{\Delta y}{\Delta x}=\frac{BE}{AB}=\cos x，即\frac{d}{dx}\sin x=\cos x 可得ΔxΔy​=ABBE​=cosx，即dxd​sinx=cosx

d d x cos ⁡ x = lim ⁡ Δ x → 0 cos ⁡ ( x + Δ x ) − cos ⁡ x Δ x = lim ⁡ Δ x → 0 cos ⁡ x ⋅ cos ⁡ Δ x − sin ⁡ x ⋅ sin ⁡ Δ x − cos ⁡ x Δ x   ( 两 角 和 公 式 cos ⁡ ( a + b ) = cos ⁡ a ⋅ cos ⁡ b − sin ⁡ a ⋅ sin ⁡ b ) = lim ⁡ Δ x → 0 cos ⁡ x ⋅ ( cos ⁡ Δ x − 1 ) Δ x − lim ⁡ Δ x → 0 sin ⁡ x ⋅ sin ⁡ Δ x Δ x = − sin ⁡ x \begin{aligned} \frac{d}{dx}\cos x&=\underset{\Delta x\rightarrow 0}{\lim}\frac{\cos \left( x+\Delta x \right) -\cos x}{\Delta x}\\ &=\underset{\Delta x\rightarrow 0}{\lim}\frac{\cos x\cdot \cos \Delta x-\sin x\cdot \sin \Delta x-\cos x}{\Delta x}\ \left(两角和公式 \cos \left( a+b \right) =\cos a\cdot \cos b-\sin a\cdot \sin b \right) \\ &=\underset{\Delta x\rightarrow 0}{\lim}\frac{\cos x\cdot \left( \cos \Delta x-1 \right)}{\Delta x}-\underset{\Delta x\rightarrow 0}{\lim}\frac{\sin x\cdot \sin \Delta x}{\Delta x}\\ &=-\sin x \end{aligned} dxd​cosx​=Δx→0lim​Δxcos(x+Δx)−cosx​=Δx→0lim​Δxcosx⋅cosΔx−sinx⋅sinΔx−cosx​ (两角和公式cos(a+b)=cosa⋅cosb−sina⋅sinb)=Δx→0lim​Δxcosx⋅(cosΔx−1)​−Δx→0lim​Δxsinx⋅sinΔx​=−sinx​

d d x tan ⁡ x = d d x sin ⁡ x cos ⁡ x = cos ⁡ 2 x + sin ⁡ 2 x cos ⁡ 2 x   ( 商的求导法则 ) = 1 cos ⁡ 2 x = sec ⁡ 2 x \begin{aligned} \frac{d}{dx}\tan x&=\frac{d}{dx}\frac{\sin x}{\cos x}\\ &=\frac{\cos ^2x+\sin ^2x}{\cos ^2x}\ \left( \text{商的求导法则} \right) \\ &=\frac{1}{\cos ^2x}\\ &=\sec ^2x \end{aligned} dxd​tanx​=dxd​cosxsinx​=cos2xcos2x+sin2x​ (商的求导法则)=cos2x1​=sec2x​

d d x csc ⁡ x = d d x ( sin ⁡ x ) − 1 = − ( sin ⁡ x ) − 2 cos ⁡ x = − cos ⁡ x sin ⁡ 2 x = − csc ⁡ x ⋅ cot ⁡ x \frac{d}{dx}\csc x=\frac{d}{dx}\left( \sin x \right) ^{-1}=-\left( \sin x \right) ^{-2}\cos x=-\frac{\cos x}{\sin ^2x}=-\csc x\cdot \cot x dxd​cscx=dxd​(sinx)−1=−(sinx)−2cosx=−sin2xcosx​=−cscx⋅cotx

d d x sec ⁡ x = d d x ( cos ⁡ x ) − 1 = − ( cos ⁡ x ) − 2 ( − sin ⁡ x ) = sin ⁡ x cos ⁡ 2 x = sec ⁡ x ⋅ tan ⁡ x \frac{d}{dx}\sec x=\frac{d}{dx}\left( \cos x \right) ^{-1}=-\left( \cos x \right) ^{-2}\left( -\sin x \right) =\frac{\sin x}{\cos ^2x}=\sec x\cdot \tan x dxd​secx=dxd​(cosx)−1=−(cosx)−2(−sinx)=cos2xsinx​=secx⋅tanx

d d x cot ⁡ x = d d x cos ⁡ x sin ⁡ x = − sin ⁡ 2 x − cos ⁡ 2 x sin ⁡ 2 x    ( 商的求导法则 ) = − 1 sin ⁡ 2 x = − csc ⁡ 2 x \begin{aligned} \frac{d}{dx}\cot x&=\frac{d}{dx}\frac{\cos x}{\sin x}\\ &=\frac{-\sin ^2x-\cos ^2x}{\sin ^2x}\,\,\left( \text{商的求导法则} \right)\\ &=-\frac{1}{\sin ^2x}\\ &=-\csc ^2x\\ \end{aligned} dxd​cotx​=dxd​sinxcosx​=sin2x−sin2x−cos2x​(商的求导法则)=−sin2x1​=−csc2x​

令  y = arcsin ⁡ x  则  sin ⁡ y = x \text{令\ }y=\arcsin x\ \text{则\ }\sin y=x 令 y=arcsinx 则 siny=x d d x ( sin ⁡ y = x ) ⇒ cos ⁡ y ⋅ d y d x = 1   \frac{d}{dx}\left( \sin y=x \right) \Rightarrow \cos y\cdot \frac{dy}{dx}=1\ dxd​(siny=x)⇒cosy⋅dxdy​=1  d y d x = 1 cos ⁡ y = 1 1 − sin ⁡ 2 y = 1 1 − x 2 \frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-\sin ^2y}}=\frac{1}{\sqrt{1-x^2}} dxdy​=cosy1​=1−sin2y ​1​=1−x2 ​1​

令  y = arccos ⁡ x  则  cos ⁡ y = x \text{令\ }y=\arccos x\ \text{则\ }\cos y=x 令 y=arccosx 则 cosy=x d d x ( cos ⁡ y = x ) ⇒ − sin ⁡ y ⋅ d y d x = 1   \frac{d}{dx}\left( \cos y=x \right) \Rightarrow -\sin y\cdot \frac{dy}{dx}=1\ dxd​(cosy=x)⇒−siny⋅dxdy​=1  d y d x = − 1 sin ⁡ y = − 1 1 − cos ⁡ 2 y = − 1 1 − x 2 \frac{dy}{dx}=-\frac{1}{\sin y}=-\frac{1}{\sqrt{1-\cos ^2y}}=-\frac{1}{\sqrt{1-x^2}} dxdy​=−siny1​=−1−cos2y ​1​=−1−x2 ​1​

令  y = arctan ⁡ x  则  tan ⁡ y = x \text{令\ }y=\arctan x\ \text{则\ }\tan y=x 令 y=arctanx 则 tany=x d d x ( tan ⁡ y = x ) ⇒ sec ⁡ 2 y ⋅ d y d x = 1   \frac{d}{dx}\left( \tan y=x \right) \Rightarrow \sec ^2y\cdot \frac{dy}{dx}=1\ dxd​(tany=x)⇒sec2y⋅dxdy​=1  d y d x = cos ⁡ 2 y \frac{dy}{dx}=\cos ^2y dxdy​=cos2y 由图可得  cos ⁡ y = 1 1 + x 2  则  d y d x = cos ⁡ 2 y = 1 1 + x 2 \text{由图可得\ }\cos y=\frac{1}{\sqrt{1+x^2}}\ \text{则\ }\frac{dy}{dx}=\cos ^2y=\frac{1}{1+x^2} 由图可得 cosy=1+x2 ​1​ 则 dxdy​=cos2y=1+x21​

sinh ⁡ ( x ) = e x − e − x 2   ,   cosh ⁡ ( x ) = e x + e − x 2   ,   tanh ⁡ ( x ) = sinh ⁡ ( x ) cosh ⁡ ( x ) = e x − e − x e x + e − x \sinh\left( x \right) =\frac{e^x-e^{-x}}{2}\ ,\ \cosh\left( x \right) =\frac{e^x+e^{-x}}{2}\ ,\ \tanh\left( x \right) =\frac{\sinh\left( x \right)}{\cosh\left( x \right)}=\frac{e^x-e^{-x}}{e^x+e^{-x}} sinh(x)=2ex−e−x​ , cosh(x)=2ex+e−x​ , tanh(x)=cosh(x)sinh(x)​=ex+e−xex−e−x​

d d x sinh ⁡ x = d d x ( e x − e − x 2 ) = e x + e − x 2 = cosh ⁡ x \frac{d}{dx}\sinh x=\frac{d}{dx}\left( \frac{e^x-e^{-x}}{2} \right) =\frac{e^x+e^{-x}}{2}=\cosh x dxd​sinhx=dxd​(2ex−e−x​)=2ex+e−x​=coshx

d d x cosh ⁡ x = d d x ( e x + e − x 2 ) = e x − e − x 2 = sinh ⁡ x \frac{d}{dx}\cosh x=\frac{d}{dx}\left( \frac{e^x+e^{-x}}{2} \right) =\frac{e^x-e^{-x}}{2}=\sinh x dxd​coshx=dxd​(2ex+e−x​)=2ex−e−x​=sinhx

d d x tanh ⁡ x = d d x sinh ⁡ x cosh ⁡ x = cosh ⁡ 2 x − sinh ⁡ 2 x cosh ⁡ 2 x    ( 商的求导法则 ) = 1 cosh ⁡ 2 x   ( 由 cosh ⁡ 2 x − sinh ⁡ 2 x = 1 可 得 ) \begin{aligned} \frac{d}{dx}\tanh x&=\frac{d}{dx}\frac{\sinh x}{\cosh x}\\ &=\frac{\cosh ^2x-\sinh ^2x}{\cosh ^2x}\,\,\left( \text{商的求导法则} \right)\\ &=\frac{1}{\cosh^2x}\ \left( 由\cosh^2x-\sinh^2x=1 可得\right)\\ \end{aligned} dxd​tanhx​=dxd​coshxsinhx​=cosh2xcosh2x−sinh2x​(商的求导法则)=cosh2x1​ (由cosh2x−sinh2x=1可得)​