一元函数求一阶导

函数求导

导数与微分

微分:对函数 y = f ( x ) y=f(x) y=f(x)定义域中的一点 x 0 x_0 x0,若存在一个只与 x 0 x_0 x0有关,而与 Δ x \Delta x Δx无关的数 g ( x 0 ) g(x_0) g(x0),使得当 Δ x → 0 \Delta x\to 0 Δx0时恒成立关系式 Δ y = g ( x 0 ) Δ x + o ( Δ x ) \Delta y=g(x_0)\Delta x+o(\Delta x) Δy=g(x0)Δx+o(Δx),则称 f ( x ) f(x) f(x) x 0 x_0 x0处可微,微分为 d y = g ( x 0 ) d x {\rm d}y=g(x_0){\rm d}x dy=g(x0)dx,即当 Δ → 0 \Delta \to 0 Δ0 Δ y \Delta y Δy的线性主部。

导数:对函数 y = f ( x ) y=f(x) y=f(x)定义域中的一点 x 0 x_0 x0,如果存在极限
lim ⁡ Δ x → 0 Δ y Δ x = lim ⁡ Δ x → 0 f ( x 0 + Δ x ) − f ( x 0 ) Δ x = d e f f ′ ( x 0 ) , \lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\stackrel {\rm def}=f'(x_0), Δx0limΔxΔy=Δx0limΔxf(x0+Δx)f(x0)=deff(x0),
则称 f ( x ) f(x) f(x) x 0 x_0 x0处可导,导数为 f ′ ( x 0 ) f'(x_0) f(x0)

命题:一元函数 f ( x ) f(x) f(x) x 0 x_0 x0处的可导与可微等价。

证明:

如果可微,即存在关系式 Δ y = g ( x 0 ) Δ x + o ( Δ x ) \Delta y=g(x_0)\Delta x+o(\Delta x) Δy=g(x0)Δx+o(Δx),则
lim ⁡ Δ x → 0 Δ y Δ x = lim ⁡ Δ x → 0 g ( x 0 ) Δ x + o ( Δ x ) Δ x = g ( x 0 ) = d e f f ′ ( x 0 ) . \lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to 0}\frac{g(x_0)\Delta x+o(\Delta x)}{\Delta x}=g(x_0)\stackrel {\rm def}=f'(x_0). Δx0limΔxΔy=Δx0limΔxg(x0)Δx+o(Δx)=g(x0)=deff(x0).
即证明了可导。

如果可导,即存在一个 f ′ ( x 0 ) f'(x_0) f(x0)使得 lim ⁡ Δ x → 0 Δ y Δ x = f ′ ( x 0 ) \lim\limits _{\Delta x\to 0}\dfrac{\Delta y}{\Delta x}=f'(x_0) Δx0limΔxΔy=f(x0),则
lim ⁡ Δ x → 0 [ Δ y Δ x − f ′ ( x 0 ) ] = lim ⁡ Δ x → 0 Δ y − f ′ ( x 0 ) Δ x Δ x = 0 , \lim_{\Delta x\to 0}[\frac{\Delta y}{\Delta x}-f'(x_0)]=\lim_{\Delta x\to 0}\frac{\Delta y-f'(x_0)\Delta x}{\Delta x}=0, Δx0lim[ΔxΔyf(x0)]=Δx0limΔxΔyf(x0)Δx=0,

Δ y − f ′ ( x 0 ) Δ x = o ( Δ x ) , Δ y = f ′ ( x 0 ) Δ x + o ( Δ x ) . \Delta y-f'(x_0)\Delta x=o(\Delta x),\quad \Delta y=f'(x_0)\Delta x+o(\Delta x). Δyf(x0)Δx=o(Δx),Δy=f(x0)Δx+o(Δx).
即证明了可微。

用定义对一些初等函数求导

常数函数的导数恒等于 0 0 0,以下出现 Δ x \Delta x Δx的式子均省略 Δ x → 0 \Delta x\to0 Δx0

一、求 y = sin ⁡ x y=\sin x y=sinx的导函数
sin ⁡ ( x + Δ x ) − sin ⁡ x Δ x = 2 cos ⁡ 2 x + Δ x 2 sin ⁡ Δ x 2 Δ x = cos ⁡ 2 x + Δ x 2 = cos ⁡ x . ⇓ ( sin ⁡ x ) ′ = cos ⁡ x . \begin{aligned} \frac{\sin (x+\Delta x)-\sin x}{\Delta x}=&\frac{2\cos \frac{2x+\Delta x}{2}\sin \frac{\Delta x}{2}}{\Delta x}\\ =&\cos \frac{2x+\Delta x}{2}\\ =&\cos x.\\ \Downarrow \\ (\sin x)'=&\cos x. \end{aligned} Δxsin(x+Δx)sinx===(sinx)=Δx2cos22x+Δxsin2Δxcos22x+Δxcosx.cosx.
二、求 y = cos ⁡ x y=\cos x y=cosx的导函数
cos ⁡ ( x + Δ x ) − cos ⁡ x Δ x = − 2 sin ⁡ 2 x + Δ x 2 sin ⁡ Δ x 2 Δ x = − sin ⁡ 2 x + Δ x 2 = − sin ⁡ x . ⇓ ( cos ⁡ x ) ′ = − sin ⁡ x . \begin{aligned} \frac{\cos (x+\Delta x)-\cos x}{\Delta x}=&\frac{-2\sin \frac{2x+\Delta x}{2}\sin \frac{\Delta x}{2}}{\Delta x}\\ =&-\sin \frac{2x+\Delta x}{2}\\ =&-\sin x.\\ \Downarrow\\ (\cos x)'=&-\sin x. \end{aligned} Δxcos(x+Δx)cosx===(cosx)=Δx2sin22x+Δxsin2Δxsin22x+Δxsinx.sinx.
三(1)、求 y = ln ⁡ x y=\ln x y=lnx的导函数
ln ⁡ ( x + Δ x ) − ln ⁡ x Δ x = ln ⁡ x + Δ x x Δ x = ln ⁡ ( 1 + Δ x x ) x ⋅ Δ x x = 1 x . ⇓ ( ln ⁡ x ) ′ = 1 x . \begin{aligned} \frac{\ln (x+\Delta x)-\ln x}{\Delta x}=&\frac{\ln\frac{x+\Delta x}{x}}{\Delta x}\\ =&\frac{\ln (1+\frac{\Delta x}{x})}{x\cdot \frac{\Delta x}{x}}\\ =&\frac 1x.\\ \Downarrow \\ (\ln x)'=&\frac 1x. \end{aligned} Δxln(x+Δx)lnx===(lnx)=Δxlnxx+ΔxxxΔxln(1+xΔx)x1.x1.
三(2)、求 y = log ⁡ a x y=\log_a x y=logax的导函数
log ⁡ a ( x + Δ x ) − log ⁡ a x Δ x = 1 ln ⁡ a ⋅ ln ⁡ ( x + Δ x ) − ln ⁡ x Δ x = 1 x ln ⁡ a . ⇓ ( log ⁡ a x ) ′ = 1 x ln ⁡ a . \begin{aligned} \frac{\log_a(x+\Delta x)-\log_a x}{\Delta x}=&\frac 1{\ln a}\cdot \frac{\ln (x+\Delta x)-\ln x}{\Delta x}\\ =&\frac {1}{x\ln a}.\\ \Downarrow \\ (\log_a x)'=&\frac 1{x\ln a}. \end{aligned} Δxloga(x+Δx)logax==(logax)=lna1Δxln(x+Δx)lnxxlna1.xlna1.
四(1)、求 y = e x y=e^x y=ex的导函数
e x + Δ x − e x Δ x = e x ⋅ e Δ x − 1 Δ x = e x . ⇓ ( e x ) ′ = e x . \begin{aligned} \frac{e^{x+\Delta x}-e^x}{\Delta x}=&e^x\cdot\frac{e^{\Delta x}-1}{\Delta x}=e^x.\\ \Downarrow \\ (e^x)'=&e^x. \end{aligned} Δxex+Δxex=(ex)=exΔxeΔx1=ex.ex.
四(2)、求 y = a x y=a^{x} y=ax的导函数
a x + Δ x − a x Δ x = a x ⋅ a Δ x − 1 Δ x = a x ln ⁡ a . ⇓ ( a x ) ′ = a x ln ⁡ a . \begin{aligned} \frac{a^{x+\Delta x}-a^x}{\Delta x}=&a^x\cdot \frac{a^{\Delta x}-1}{\Delta x}\\ =&a^x\ln a.\\ \Downarrow \\ (a^x)'=&a^x\ln a. \end{aligned} Δxax+Δxax==(ax)=axΔxaΔx1axlna.axlna.
五、求 y = x a y=x^a y=xa的导函数
( x + Δ x ) a − x a Δ x = x a [ ( 1 + Δ x x ) a − 1 ] x ⋅ Δ x x = a x a − 1 . ⇓ ( x a ) ′ = a x a − 1 . \begin{aligned} \frac{(x+\Delta x)^a-x^a}{\Delta x}=&\frac{x^a[(1+\frac{\Delta x}{x})^a-1]}{x\cdot\frac{\Delta x}{x}}\\ =&ax^{a-1}.\\ \Downarrow \\ (x^a)'=&ax^{a-1}. \end{aligned} Δx(x+Δx)axa==(xa)=xxΔxxa[(1+xΔx)a1]axa1.axa1.

求导(微)的四则运算法则

法则1(线性求导): [ c 1 f ( x ) + c 2 g ( x ) ] ′ = c 1 f ′ ( x ) + c 2 g ′ ( x ) [c_1f(x)+c_2g(x)]'=c_1f'(x)+c_2g'(x) [c1f(x)+c2g(x)]=c1f(x)+c2g(x) d [ c 1 f + c 2 g ] = c 1 d f + c 2 d g {\rm d}[c_1f+c_2g]=c_1{\rm d}f+c_2{\rm d}g d[c1f+c2g]=c1df+c2dg

证明:
[ c 1 f ( x ) + c 2 g ( x ) ] ′ = [ c 1 f ( x + Δ x ) + c 2 g ( x + Δ x ) ] − [ c 1 f ( x ) + c 2 g ( x ) ] Δ x = c 1 [ f ( x + Δ x ) − f ( x ) ] Δ x + c 2 [ g ( x + Δ x ) − g ( x ) ] Δ x = c 1 f ′ ( x ) + c 2 g ′ ( x ) . \begin{aligned} [c_1f(x)+c_2g(x)]'=&\frac{[c_1f(x+\Delta x)+c_2g(x+\Delta x)]-[c_1f(x)+c_2g(x)]}{\Delta x}\\ =&\frac{c_1[f(x+\Delta x)-f(x)]}{\Delta x}+\frac{c_2[g(x+\Delta x)-g(x)]}{\Delta x}\\ =&c_1f'(x)+c_2g'(x). \end{aligned} [c1f(x)+c2g(x)]===Δx[c1f(x+Δx)+c2g(x+Δx)][c1f(x)+c2g(x)]Δxc1[f(x+Δx)f(x)]+Δxc2[g(x+Δx)g(x)]c1f(x)+c2g(x).
法则2(乘法求导): [ f ( x ) ⋅ g ( x ) ] ′ = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) [f(x)\cdot g(x)]'=f'(x)g(x)+f(x)g'(x) [f(x)g(x)]=f(x)g(x)+f(x)g(x) d [ f g ] = ( d f ) g + f ( d g ) {\rm d}[fg]=({\rm d}f)g+f({\rm d}g) d[fg]=(df)g+f(dg)

证明:
[ f ( x ) g ( x ) ] ′ = f ( x + Δ x ) g ( x + Δ x ) − f ( x ) g ( x ) Δ x = f ( x + Δ x ) g ( x + Δ x ) − f ( x + Δ x ) g ( x ) Δ x + f ( x + Δ x ) g ( x ) − f ( x ) g ( x ) Δ x = f ( x + Δ x ) g ′ ( x ) + g ( x ) f ′ ( x ) = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) . \begin{aligned} [f(x)g(x)]'=&\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}\\ =&\frac{f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x)}{\Delta x}+\frac{f(x+\Delta x)g(x)-f(x)g(x)}{\Delta x}\\ =&f(x+\Delta x)g'(x)+g(x)f'(x)\\ =&f'(x)g(x)+f(x)g'(x). \end{aligned} [f(x)g(x)]====Δxf(x+Δx)g(x+Δx)f(x)g(x)Δxf(x+Δx)g(x+Δx)f(x+Δx)g(x)+Δxf(x+Δx)g(x)f(x)g(x)f(x+Δx)g(x)+g(x)f(x)f(x)g(x)+f(x)g(x).
法则3(倒数求导): [ 1 g ( x ) ] ′ = − g ′ ( x ) [ g ( x ) ] 2 [\dfrac 1{g(x)}]'=-\dfrac{g'(x)}{[g(x)]^2} [g(x)1]=[g(x)]2g(x) d ( 1 g ) = − d g g 2 {\rm d}(\dfrac{1}{g})=-\dfrac{ {\rm d}g}{g^2} d(g1)=g2dg

证明:
[ 1 g ( x ) ] ′ = 1 g ( x + Δ x ) − 1 g ( x ) Δ x = 1 g ( x + Δ x ) g ( x ) ⋅ g ( x ) − g ( x + Δ x ) Δ x = − g ′ ( x ) [ g ( x ) ] 2 . \begin{aligned} \left[\frac 1{g(x)}\right]'=&\frac{\frac 1{g(x+\Delta x)}-\frac 1{g(x)}}{\Delta x}\\ =&\frac{1}{g(x+\Delta x)g(x)}\cdot\frac{g(x)-g(x+\Delta x)}{\Delta x}\\ =&\frac{-g'(x)}{[g(x)]^2}. \end{aligned} [g(x)1]===Δxg(x+Δx)1g(x)1g(x+Δx)g(x)1Δxg(x)g(x+Δx)[g(x)]2g(x).
法则4(除法求导): [ f ( x ) g ( x ) ] ′ = f ′ ( x ) g ( x ) − f ( x ) g ( x ) [ g ( x ) ] 2 [\dfrac {f(x)}{g(x)}]'=\dfrac{f'(x)g(x)-f(x)g(x)}{[g(x)]^2} [g(x)f(x)]=[g(x)]2f(x)g(x)f(x)g(x) d [ f g ] = ( d f ) g − f ( d g ) g 2 {\rm d}[\dfrac fg]=\dfrac{({\rm d}f)g-f({\rm d}g)}{g^2} d[gf]=g2(df)gf(dg)

证明:法则2+法则3=法则4。

法则5(多项乘法求导):可用数学归纳法证明。
[ ∏ i = 1 n f i ( x ) ] ′ = ∑ i = 1 n [ f i ′ ( x ) ∏ j = 1 ; j ≠ i n f j ( x ) ] . \left[\prod_{i=1}^n f_i(x)\right]'=\sum_{i=1}^n \left[f_i'(x)\prod_{j=1;j\ne i}^nf_j(x)\right]. [i=1nfi(x)]=i=1nfi(x)j=1;j=infj(x).

反函数求导

命题:若函数 y = f ( x ) y=f(x) y=f(x) ( a , b ) (a,b) (a,b)上连续、严格单调、可导且 f ′ ( x ) ≠ 0 f'(x)\ne 0 f(x)=0,记 α = min ⁡ ( f ( a + ) , f ( b − ) ) , β = max ⁡ ( f ( a + ) , f ( b − ) ) \alpha=\min (f(a^+),f(b^-)),\beta=\max (f(a^+),f(b^-)) α=min(f(a+),f(b)),β=max(f(a+),f(b)),则反函数 x = f − 1 ( y ) x=f^{-1}(y) x=f1(y) ( α , β ) (\alpha,\beta) (α,β)上可导,且 [ f − 1 ( y ) ] ′ = 1 f ′ ( x ) [f^{-1}(y)]'=\dfrac1{f'(x)} [f1(y)]=f(x)1

证明:

由反函数存在定理, x = f − 1 ( y ) x=f^{-1}(y) x=f1(y)存在,且 Δ y = f ( x + Δ x ) − f ( x ) ≠ 0 \Delta y=f(x+\Delta x)-f(x)\ne 0 Δy=f(x+Δx)f(x)=0,等价于
Δ x = f − 1 ( y + Δ y ) − f − 1 ( y ) ≠ 0. \Delta x=f^{-1}(y+\Delta y)-f^{-1}(y)\ne 0. Δx=f1(y+Δy)f1(y)=0.
且当 Δ x → 0 \Delta x\to 0 Δx0时有 Δ y → 0 \Delta y\to 0 Δy0(以上可以通过图像看出)。
[ f − 1 ( y ) ] ′ = lim ⁡ Δ y → 0 f − 1 ( y + Δ y ) − f − 1 ( y ) Δ y = lim ⁡ Δ x → 0 Δ x f ( x + Δ x ) − f ( x ) = 1 f ′ ( x ) . \begin{aligned} [f^{-1}(y)]'=&\lim_{\Delta y\to 0}\frac{f^{-1}(y+\Delta y)-f^{-1}(y)}{\Delta y}\\ =&\lim_{\Delta x\to 0}\frac{\Delta x}{f(x+\Delta x)-f(x)}\\ =&\frac{1}{f'(x)}. \end{aligned} [f1(y)]===Δy0limΔyf1(y+Δy)f1(y)Δx0limf(x+Δx)f(x)Δxf(x)1.
这一点可以记作
d x d y = [ d y d x ] − 1 . \frac{ {\rm d} x}{ {\rm d} y}=\left[\frac{ {\rm d} y}{ {\rm d} x}\right]^{-1}. dydx=[dxdy]1.

用反函数求导求反三角函数的导数

一、求 y = arcsin ⁡ x , x ∈ [ − 1 , 1 ] y=\arcsin x,x\in [-1,1] y=arcsinx,x[1,1]的导函数

这里 x = sin ⁡ y , y = [ − π 2 , π 2 ] x=\sin y,y=[-\frac \pi2,\frac \pi2] x=siny,y=[2π,2π]是其导函数,所以
d y d x = [ d x d y ] − 1 = 1 cos ⁡ y = 1 cos ⁡ arcsin ⁡ x = 1 1 − x 2 . \begin{aligned} &\frac{ {\rm d}y}{ {\rm d}x}=\left[\frac{ {\rm d}x}{ {\rm d}y}\right]^{-1}\\ =&\frac{1}{\cos y}\\ =&\frac{1}{\cos \arcsin x}\\ =&\frac{1}{\sqrt{1-x^2}}. \end{aligned} ===dxdy=[dydx]1cosy1cosarcsinx11x2 1.
二、求 y = arccos ⁡ x , x ∈ [ − 1 , 1 ] y=\arccos x,x\in [-1,1] y=arccosx,x[1,1]的导函数

这里 x = cos ⁡ y , y ∈ [ 0 , π ] x=\cos y,y\in [0,\pi] x=cosy,y[0,π]是其反函数,所以
d y d x = [ d x d y ] − 1 = − 1 sin ⁡ y = − 1 1 − cos ⁡ 2 y = − 1 1 − x 2 . \begin{aligned} &\frac{ {\rm d}y}{ {\rm d}x}=\left[\frac{ {\rm d}x}{ {\rm d}y}\right]^{-1}\\ =&-\frac{1}{\sin y}\\ =&-\frac{1}{\sqrt{1-\cos^2 y}}\\ =&-\frac{1}{\sqrt{1-x^2}}. \end{aligned} ===dxdy=[dydx]1siny11cos2y 11x2 1.
三、求 y = arctan ⁡ x , x ∈ R y=\arctan x,x\in \R y=arctanx,xR的导函数

这里 x = tan ⁡ y , y ∈ [ − π 2 , π 2 ] x=\tan y,y\in [-\frac \pi 2,\frac \pi2] x=tany,y[2π,2π]是其反函数,所以
d y d x = [ d x d y ] − 1 = 1 sec ⁡ 2 y = 1 1 + tan ⁡ 2 y = 1 1 + x 2 . \begin{aligned} &\frac{ {\rm d}y}{ {\rm d}x}=\left[\frac{ {\rm d}x}{ {\rm d}y}\right]^{-1}\\ =&\frac{1}{\sec^2 y}\\ =&\frac{1}{1+\tan^2y}\\ =&\frac{1}{1+x^2}. \end{aligned} ===dxdy=[dydx]1sec2y11+tan2y11+x21.

复合函数求导的链式法则

命题:设 u = g ( x ) u=g(x) u=g(x) x = x 0 x=x_0 x=x0处可导且 y = f ( u ) y=f(u) y=f(u) u = u 0 = g ( x 0 ) u=u_0=g(x_0) u=u0=g(x0)处可导,则复合函数 y = f ( g ( x ) ) y=f(g(x)) y=f(g(x)) x = x 0 x=x_0 x=x0处可导,且有 [ f ( g ( x ) ) ] ′ = f ′ ( u 0 ) g ′ ( x 0 ) = f ′ ( g ( x 0 ) ) g ′ ( x 0 ) [f(g(x))]'=f'(u_0)g'(x_0)=f'(g(x_0))g'(x_0) [f(g(x))]=f(u0)g(x0)=f(g(x0))g(x0)

证明:

由于 y = f ( u ) y=f(u) y=f(u) u 0 u_0 u0处可导,所以可微,故存在一个无穷小量 α \alpha α满足 lim ⁡ Δ u → 0 α = 0 \lim\limits_{\Delta u\to 0}\alpha=0 Δu0limα=0,且
f ( u 0 + Δ u ) − f ( u 0 ) = f ′ ( u 0 ) Δ u + α Δ u . f(u_0+\Delta u)-f(u_0)=f'(u_0)\Delta u+\alpha\Delta u. f(u0+Δu)f(u0)=f(u0)Δu+αΔu.
如果让 Δ u = 0 \Delta u=0 Δu=0 α = 0 \alpha=0 α=0,则上式对 Δ u = 0 \Delta u=0 Δu=0依然成立。设 Δ u = g ( x 0 + Δ x ) − g ( x 0 ) \Delta u=g(x_0+\Delta x)-g(x_0) Δu=g(x0+Δx)g(x0),这样就有
f ( u 0 + Δ u ) − f ( u 0 ) Δ x = f ( g ( x 0 + Δ x ) ) − f ( g ( x 0 ) ) Δ x = f ′ ( u 0 ) Δ u Δ x + α Δ u Δ x . \frac{f(u_0+\Delta u)-f(u_0)}{\Delta x}=\frac{f(g(x_0+\Delta x))-f(g(x_0))}{\Delta x}=f'(u_0)\frac{\Delta u}{\Delta x}+\alpha\frac{\Delta u}{\Delta x}. Δxf(u0+Δu)f(u0)=Δxf(g(x0+Δx))f(g(x0))=f(u0)ΔxΔu+αΔxΔu.
Δ x → 0 \Delta x\to 0 Δx0,则由 Δ u = g ( u 0 ) Δ x + o ( Δ x ) \Delta u=g(u_0)\Delta x+o(\Delta x) Δu=g(u0)Δx+o(Δx) Δ u → 0 \Delta u\to 0 Δu0,于是
lim ⁡ Δ x → 0 f ( g ( x 0 + Δ x ) ) − f ( g ( x 0 ) ) Δ x = lim ⁡ Δ x → 0 f ′ ( u 0 ) g ′ ( x 0 ) + g ′ ( x 0 ) lim ⁡ Δ u → 0 α = f ′ ( u 0 ) g ′ ( x 0 ) . \lim_{\Delta x\to 0}\frac{f(g(x_0+\Delta x))-f(g(x_0))}{\Delta x}=\lim_{\Delta x\to 0}f'(u_0)g'(x_0)+g'(x_0)\lim_{\Delta u\to 0}\alpha=f'(u_0)g'(x_0). Δx0limΔxf(g(x0+Δx))f(g(x0))=Δx0limf(u0)g(x0)+g(x0)Δu0limα=f(u0)g(x0).
也可以写成链式法则:
d y d x = d y d u ⋅ d u d x . \frac{ {\rm d}y}{ {\rm d}x}=\frac{ {\rm d}y}{ {\rm d}u}\cdot \frac{ {\rm d}u}{ {\rm d}x}. dxdy=dudydxdu.
微分形式为
d [ f ( g ( x ) ) ] = f ′ ( g ( x ) ) g ′ ( x ) d x . {\rm d}[f(g(x))]=f'(g(x))g'(x){\rm d}x. d[f(g(x))]=f(g(x))g(x)dx.

由此可以推出一阶微分的形式不变性,即 u = g ( x ) , y = f ( u ) u=g(x),y=f(u) u=g(x),y=f(u),有
d [ f ( u ) ] = f ′ ( u ) d u . {\rm d}[f(u)]=f'(u){\rm d}u. d[f(u)]=f(u)du.
虽然这里 u u u不是自变量,但依然可以把 u u u当成自变量来计算 f ( u ) f(u) f(u)的微分。

对数求导法

对数求导法适用于形如 y = f ( x ) = u ( x ) v ( x ) y=f(x)=u(x)^{v(x)} y=f(x)=u(x)v(x)的函数求导,这种函数也叫幂指函数。

首先令 z ( x ) = ln ⁡ y = v ( x ) ln ⁡ u ( x ) z(x)=\ln y=v(x)\ln u(x) z(x)=lny=v(x)lnu(x),则
z ′ ( x ) = v ′ ( x ) ln ⁡ u ( x ) + v ( x ) u ′ ( x ) u ( x ) , z ′ ( x ) = y ′ ( x ) y ( x ) , z'(x)=v'(x)\ln u(x)+\frac{v(x)u'(x)}{u(x)},\\ z'(x)=\frac{y'(x)}{y(x)}, z(x)=v(x)lnu(x)+u(x)v(x)u(x),z(x)=y(x)y(x),
综合两式得到
y ′ ( x ) = [ v ′ ( x ) ln ⁡ u ( x ) + v ( x ) u ′ ( x ) u ( x ) ] ⋅ u ( x ) v ( x ) . y'(x)=\left[v'(x)\ln u(x)+\frac{v(x)u'(x)}{u(x)} \right]\cdot u(x)^{v(x)}. y(x)=[v(x)lnu(x)+u(x)v(x)u(x)]u(x)v(x).
这是一种将复杂函数转化为简单初等函数求导的方法,重点是找到一个转换关系充当 y y y u , v u,v u,v之间的桥梁。

隐函数求导

隐函数求导依赖于复合函数的求导法则、乘法求导法则与一阶微分的形式不变性。在求导时直接对隐函数的两边同时求微分或者求导即可。如
e x y + x 2 y = 1 , e^{xy}+x^2y=1, exy+x2y=1,
两边同时求导,有
( y + x y ′ ) e x y + 2 x y + x 2 y ′ = 0 (y+xy')e^{xy}+2xy+x^2y'=0 (y+xy)exy+2xy+x2y=0
解出
y ′ = − y e x y + 2 x y x e x y + x 2 = − y ( e x y + 2 x ) x ( e x y + x ) . y'=-\frac{ye^{xy}+2xy}{xe^{xy}+x^2}=-\frac{y(e^{xy}+2x)}{x(e^{xy}+x)}. y=xexy+x2yexy+2xy=x(exy+x)y(exy+2x).
对于变量分离的隐函数 g ( y ) = f ( x ) g(y)=f(x) g(y)=f(x),两边同时求微分,得到
g ′ ( y ) d y = f ′ ( x ) d x , d y d x = f ′ ( x ) g ′ ( y ) . g'(y){\rm d}y=f'(x){\rm d}x,\quad \frac{ {\rm d}y}{ {\rm d}x}=\frac{f'(x)}{g'(y)}. g(y)dy=f(x)dx,dxdy=g(y)f(x).
这里 g ′ ( y ) g'(y) g(y)是将 y y y视为自变量,对 y y y的导数。

隐函数求导得到的导数一般含有 y y y,但在使用时没有妨碍。

参数方程求导

x , y x,y x,y的函数关系由参数形式确定,即
{ x = φ ( t ) , y = ψ ( t ) . \left\{ \begin{array}l x=\varphi(t),\\ y=\psi(t). \end{array} \right. { x=φ(t),y=ψ(t).

如果 φ ( t ) \varphi(t) φ(t)严格单调且 φ ( t ) ≠ 0 \varphi(t)\ne0 φ(t)=0,则有
d y d x = d y / d t d x / d t = ψ ′ ( t ) φ ′ ( t ) = ψ ′ ( φ − 1 ( x ) ) φ ′ ( φ − 1 ( x ) ) . \frac{ {\rm d}y}{ {\rm d}x}=\frac{ {\rm d}y/{\rm d}t}{ {\rm d}x/{\rm d}t}=\frac{\psi'(t)}{\varphi'(t)}=\frac{\psi'(\varphi^{-1}(x))}{\varphi'(\varphi^{-1}(x))}. dxdy=dx/dtdy/dt=φ(t)ψ(t)=φ(φ1(x))ψ(φ1(x)).
证明:

因为 x = φ ( t ) x=\varphi(t) x=φ(t)严格单调且递增,所以存在反函数 t = φ − 1 ( x ) t=\varphi^{-1}(x) t=φ1(x),所以 y = ψ ( φ − 1 ( x ) ) y=\psi(\varphi^{-1}(x)) y=ψ(φ1(x)),由复合函数求导法则得到
d y d x = ψ ′ ( φ − 1 ( x ) ) ⋅ [ φ − 1 ( x ) ] ′ = ψ ′ ( t ) φ ′ ( t ) . \frac{ {\rm d}y}{ {\rm d}x}=\psi'(\varphi^{-1}(x))\cdot [\varphi^{-1}(x)]'=\frac{\psi'(t)}{\varphi'(t)}. dxdy=ψ(φ1(x))[φ1(x)]=φ(t)ψ(t).

你可能感兴趣的:(数学分析)