4. 微分
4.4 复合函数求导法则及其应用
4.4.1 复合函数求导法则
【定理4.4.1】 u = g ( x ) u=g(x) u=g(x)在 x = x 0 x=x_0 x=x0可导， g ( x 0 ) = u ( 0 ) g(x_0)=u(0) g(x0)=u(0)， y = f ( u ) y=f(u) y=f(u)在 u = u 0 u=u_0 u=u0可导，则 y = f ( g ( x ) ) y=f(g(x)) y=f(g(x))在 x = x 0 x=x_0 x=x0可导且 [ f ( g ( x ) ) ] x = x 0 ′ = f ′ ( u 0 ) g ′ ( x 0 ) [f(g(x))]'_{x=x_0}=f'(u_0)g'(x_0) [f(g(x))]x=x0′=f′(u0)g′(x0)
【一个有缺陷的证明 ！！！】 g ( x 0 + Δ x ) − g ( x 0 ) = Δ u g(x_0+\Delta x)-g(x_0)=\Delta u g(x0+Δx)−g(x0)=Δu， Δ y = f ( u 0 + Δ u ) − f ( u 0 ) \Delta y=f(u_0+\Delta u)-f(u_0) Δy=f(u0+Δu)−f(u0)， [ f ( g ( x ) ) ] x = x 0 ′ = lim Δ x → 0 Δ y Δ x [f(g(x))]'{x=x_0}=\lim\limits{\Delta x\to 0}\frac{\Delta y}{\Delta x} [f(g(x))]x=x0′=Δx→0limΔxΔy
当 Δ → 0 , Δ u → 0 \Delta \to 0,\Delta u \to 0 Δ→0,Δu→0，则：
[ f ( g ( x ) ) ] x = x 0 ′ = lim Δ x → 0 Δ y Δ x = lim Δ x → 0 ( Δ y Δ u ⋅ Δ u Δ x ) = lim Δ u → 0 Δ y Δ u ⋅ lim Δ x → 0 Δ u Δ x = f ′ ( u 0 ) g ′ ( x 0 ) [f(g(x))]'{x=x_0}=\lim\limits{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\to 0}(\frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x})=\lim\limits_{\Delta u\to 0}\frac{\Delta y}{\Delta u}\cdot\lim\limits_{\Delta x\to 0}\frac{\Delta u}{\Delta x}=f'(u_0)g'(x_0) [f(g(x))]x=x0′=Δx→0limΔxΔy=Δx→0lim(ΔuΔy⋅ΔxΔu)=Δu→0limΔuΔy⋅Δx→0limΔxΔu=f′(u0)g′(x0)
【问题所在】 Δ u → 0 , Δ u ≠ 0 \Delta u\to 0,\Delta u\ne 0 Δu→0,Δu=0，但是 lim Δ u → 0 Δ y Δ u \lim\limits_{\Delta u\to 0}\frac{\Delta y}{\Delta u} Δu→0limΔuΔy中， Δ u \Delta u Δu可能为0，比如有下面的反例：
u ( x ) = { x 2 sin 1 x 0 u(x)=\left\{\begin{array}{c} x^{2} \sin \frac{1}{x} \\ 0 \end{array}\right. u(x)={x2sinx10，如果求0点的复合函数导数，则
Δ u = u ( 0 + Δ x ) − u ( 0 ) = ( Δ x ) 2 sin 1 Δ x \Delta u =u(0+\Delta x)-u(0)=(\Delta x)^2\sin\frac{1}{\Delta x} Δu=u(0+Δx)−u(0)=(Δx)2sinΔx1
则 u ′ ( 0 ) = lim Δ x → 0 ( Δ x ) 2 sin 1 Δ x Δ x = 0 u'(0)=\lim\limits_{\Delta x\to 0}\frac{(\Delta x)^2\sin\frac{1}{\Delta x}}{\Delta x}=0 u′(0)=Δx→0limΔx(Δx)2sinΔx1=0，根据海涅定理，取 Δ x n = 1 n π ≠ 0 \Delta x_n=\frac{1}{n\pi}\ne 0 Δxn=nπ1=0，但是此时 Δ u = 1 n 2 π 2 sin ( n π ) = 0 \Delta u=\frac{1}{n^2\pi^2}\sin(n\pi)=0 Δu=n2π21sin(nπ)=0，这就出现了 Δ u = 0 \Delta u = 0 Δu=0的情况，如果这个函数和别的函数 f ( u ) f(u) f(u)复合，则在计算 lim Δ u → 0 Δ y Δ u \lim\limits_{\Delta u\to 0}\frac{\Delta y}{\Delta u} Δu→0limΔuΔy时，会出现分母确确实实是真0而不是趋近于0的情况，所以不能这样证明。
【更改】由 f ( u ) f(u) f(u)可导，则 f ( u ) f(u) f(u)可微， Δ y = f ( u 0 + Δ u ) − f ( u 0 ) = f ′ ( u 0 ) Δ u + o ( Δ u ) \Delta y=f(u_0+\Delta u)-f(u_0)=f'(u_0)\Delta u+o(\Delta u) Δy=f(u0+Δu)−f(u0)=f′(u0)Δu+o(Δu)
令 α = o ( Δ u ) Δ u , lim Δ u → 0 , Δ u ≠ 0 α = 0 \alpha = \frac{o(\Delta u)}{\Delta u},\lim\limits_{\Delta u\to 0,\Delta u \ne 0}\alpha = 0 α=Δuo(Δu),Δu→0,Δu=0limα=0
则 Δ y = f ( u 0 + Δ u ) − f ( u 0 ) = f ′ ( u 0 ) Δ u + α Δ u ( Δ u → 0 ) \Delta y=f(u_0+\Delta u)-f(u_0)=f'(u_0)\Delta u+\alpha \Delta u(\Delta u \to 0) Δy=f(u0+Δu)−f(u0)=f′(u0)Δu+αΔu(Δu→0)
当 Δ u = 0 , α = 0 \Delta u = 0,\alpha = 0 Δu=0,α=0，对上式 Δ y = f ′ ( u 0 ) Δ u + α Δ u \Delta y=f'(u_0)\Delta u+\alpha\Delta u Δy=f′(u0)Δu+αΔu也成立
Δ y Δ x = f ′ ( u 0 ) Δ u Δ x + α Δ u Δ x \frac{\Delta y}{\Delta x}=f'(u_0)\frac{\Delta u}{\Delta x}+\alpha\frac{\Delta u}{\Delta x} ΔxΔy=f′(u0)ΔxΔu+αΔxΔu
令 Δ x → 0 ⇒ Δ u → 0 ( Δ u = 0 或 Δ u ≠ 0 ) ⇒ lim Δ u → 0 α = 0 \Delta x\to 0\Rightarrow\Delta u\to 0(\Delta u = 0或\Delta u \ne 0)\Rightarrow\lim\limits_{\Delta u\to 0}\alpha = 0 Δx→0⇒Δu→0(Δu=0或Δu=0)⇒Δu→0limα=0
lim Δ x → 0 Δ y Δ x = lim Δ x → 0 f ′ ( u 0 ) Δ u Δ x + lim Δ x → 0 α Δ u Δ x = f ′ ( u 0 ) g ′ ( x 0 ) + 0 × g ′ ( x 0 ) = f ′ ( u 0 ) g ′ ( x 0 ) \lim\limits_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\to 0}f'(u_0)\frac{\Delta u}{\Delta x}+\lim\limits_{\Delta x\to 0}\alpha\frac{\Delta u}{\Delta x}=f'(u_0)g'(x_0)+0\times g'(x_0)=f'(u_0)g'(x_0) Δx→0limΔxΔy=Δx→0limf′(u0)ΔxΔu+Δx→0limαΔxΔu=f′(u0)g′(x0)+0×g′(x0)=f′(u0)g′(x0)
【复合函数链式求导法则】
y = f ( u ) , u = g ( x ) , y = f ( g ( x ) ) = f ∘ g ( x ) y=f(u),u=g(x),y=f(g(x))=f\circ g(x) y=f(u),u=g(x),y=f(g(x))=f∘g(x)
[ f ( g ( x ) ) ] ′ = d y d x = d y d u ⋅ d u d x [f(g(x))]'=\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} [f(g(x))]′=dxdy=dudy⋅dxdu
【例】 y = f ( u ) , u = g ( x ) , x = h ( t ) , d y d t = d y d u ⋅ d u d x ⋅ d x d t = f ′ ( u 0 ) g ′ ( x 0 ) h ′ ( t 0 ) y=f(u),u=g(x),x=h(t),\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dx}\cdot\frac{dx}{dt}=f'(u_0)g'(x_0)h'(t_0) y=f(u),u=g(x),x=h(t),dtdy=dudy⋅dxdu⋅dtdx=f′(u0)g′(x0)h′(t0)
【例4.4.1】 y = x α = e α ln x , y = e u , u = α ln x , d y d x = ( e u ) ′ ( α ln x ) ′ = e u ⋅ α x = x α ⋅ α x = α x α − 1 y=x^{\alpha}=e^{\alpha \ln x},y=e^u,u=\alpha \ln x,\frac{dy}{dx}=(e^u)'(\alpha \ln x)'=e^u\cdot\frac{\alpha}{x}=x^{\alpha}\cdot\frac{\alpha}{x}=\alpha x^{\alpha - 1} y=xα=eαlnx,y=eu,u=αlnx,dxdy=(eu)′(αlnx)′=eu⋅xα=xα⋅xα=αxα−1
【例4.4.2】 y = e cos x y=e^{\cos x} y=ecosx，求 y ′ y' y′
【解】 d y d x = ( e u ) ′ ( cos x ) ′ = − e cos x sin x \frac{dy}{dx}=(e^u)'(\cos x)'=-e^{\cos x}\sin x dxdy=(eu)′(cosx)′=−ecosxsinx
【例】求 ( 1 + x 2 ) ′ (\sqrt{1+x^2})' (1+x2 )′
【解】 ( 1 + x 2 ) ′ = 1 2 1 + x 2 ⋅ 2 x = x 1 + x 2 (\sqrt{1+x^2})'=\frac{1}{2\sqrt{1+x^2}}\cdot2x=\frac{x}{\sqrt{1+x^2}} (1+x2 )′=21+x2 1⋅2x=1+x2 x