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 Δx→0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)在 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 Δ→0时 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), Δx→0limΔxΔy=Δx→0limΔxf(x0+Δx)−f(x0)=deff′(x0),
则称 f ( x ) f(x) f(x)在 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)在 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). Δx→0limΔxΔy=Δx→0limΔxg(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) Δx→0limΔxΔy=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, Δx→0lim[ΔxΔy−f′(x0)]=Δx→0limΔxΔy−f′(x0)Δx=0,
即
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). Δy−f′(x0)Δx=o(Δx),Δy=f′(x0)Δx+o(Δx).
即证明了可微。
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 Δx→0。
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)′=Δx−2sin22x+Δxsin2Δx−sin22x+Δx−sinx.−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+Δxx⋅xΔxln(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)′=lna1⋅Δxln(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+Δx−ex=⇓(ex)′=ex⋅ΔxeΔx−1=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+Δx−ax==⇓(ax)′=ax⋅ΔxaΔx−1axlna.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)a−xa==⇓(xa)′=x⋅xΔxxa[(1+xΔx)a−1]axa−1.axa−1.
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)1−g(x)1g(x+Δx)g(x)1⋅Δxg(x)−g(x+Δx)[g(x)]2−g′(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)g−f(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=1∏nfi(x)]′=i=1∑n⎣⎡fi′(x)j=1;j=i∏nfj(x)⎦⎤.
命题:若函数 y = f ( x ) y=f(x) y=f(x)在 ( 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=f−1(y)在 ( α , β ) (\alpha,\beta) (α,β)上可导,且 [ f − 1 ( y ) ] ′ = 1 f ′ ( x ) [f^{-1}(y)]'=\dfrac1{f'(x)} [f−1(y)]′=f′(x)1。
证明:
由反函数存在定理, x = f − 1 ( y ) x=f^{-1}(y) x=f−1(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=f−1(y+Δy)−f−1(y)=0.
且当 Δ x → 0 \Delta x\to 0 Δx→0时有 Δ y → 0 \Delta y\to 0 Δy→0(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} [f−1(y)]′===Δy→0limΔyf−1(y+Δy)−f−1(y)Δx→0limf(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.
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]−1cosy1cosarcsinx11−x21.
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]−1−siny1−1−cos2y1−1−x21.
三、求 y = arctan x , x ∈ R y=\arctan x,x\in \R y=arctanx,x∈R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.
命题:设 u = g ( x ) u=g(x) u=g(x)在 x = x 0 x=x_0 x=x0处可导且 y = f ( u ) y=f(u) y=f(u)在 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))在 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)在 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 Δu→0limα=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时 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.
令 Δ x → 0 \Delta x\to 0 Δx→0,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)得 Δ u → 0 \Delta u\to 0 Δu→0,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). Δx→0limΔxf(g(x0+Δx))−f(g(x0))=Δx→0limf′(u0)g′(x0)+g′(x0)Δu→0limα=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=dudy⋅dxdu.
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与 u , v u,v u,v之间的桥梁。
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。
设 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′(φ−1(x))ψ′(φ−1(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=φ−1(x),所以 y = ψ ( φ − 1 ( x ) ) y=\psi(\varphi^{-1}(x)) y=ψ(φ−1(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′(φ−1(x))⋅[φ−1(x)]′=φ′(t)ψ′(t).