4.5 高阶导数和高阶微分
隐函数求高阶导数：
【例4.5.7】 e x y + x 2 y − 1 = 0 e^{xy}+x^2y-1=0 exy+x2y−1=0，求 d 2 y d x 2 \frac{d^2y}{dx^2} dx2d2y.
【解】在等式两侧关于 x x x求导得
e x y ( y + x y ′ ) + 2 x y + x 2 y ′ = 0 e^{xy}(y+xy')+2xy+x^2y'=0 exy(y+xy′)+2xy+x2y′=0…(1)
整理得 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+x2−(yexy+2xy)=−x(exy+x)y(exy+2x)
从(1)式出发继续求导会简单一些
对(1)式两侧关于 x x x求导得
e x y ( y + x y ′ ) 2 + e x y ( y ′ + y ′ + x y ′ ′ ) + 2 y + 2 x y ′ + 2 x y ′ + x 2 y ′ ′ = 0 e^{xy}(y+xy')^2+e^{xy}(y'+y'+xy'')+2y+2xy'+2xy'+x^2y''=0 exy(y+xy′)2+exy(y′+y′+xy′′)+2y+2xy′+2xy′+x2y′′=0
即 e x y ( y + x y ′ ) 2 + e x y ( 2 y ′ + x y ′ ′ ) + 2 y + 4 x y ′ + x 2 y ′ ′ = 0 e^{xy}(y+xy')^2+e^{xy}(2y'+xy'')+2y+4xy'+x^2y''=0 exy(y+xy′)2+exy(2y′+xy′′)+2y+4xy′+x2y′′=0
整理得
y ′ = e x y ( ( y + x y ′ ) 2 + 2 y ′ ) + ( 2 y + 4 x y ′ ) x ( e x y + x ) = 2 y e 3 x y + 8 x y e 2 x y + ( 12 x 2 y − x 3 y 2 ) e x y + 6 x 3 y x 2 ( e x y + x ) 3 y'=\frac{e^{xy}((y+xy')^2+2y')+(2y+4xy')}{x(e^{xy}+x)}=\frac{2 y \mathrm{e}^{3 x y}+8 x y \mathrm{e}^{2 x y}+\left(12 x^{2} y-x^{3} y^{2}\right) \mathrm{e}^{x y}+6 x^{3} y}{x^{2}\left(\mathrm{e}^{x y}+x\right)^{3}} y′=x(exy+x)exy((y+xy′)2+2y′)+(2y+4xy′)=x2(exy+x)32ye3xy+8xye2xy+(12x2y−x3y2)exy+6x3y
将 y ′ y' y′代入上式
参数表示的函数的高阶导数
{ x = φ ( t ) , y = ψ ( t ) , α ⩽ t ⩽ β \left\{\begin{array}{l} x=\varphi(t), \\ y=\psi(t), \end{array} \quad \alpha \leqslant t \leqslant \beta\right. {x=φ(t),y=ψ(t),α⩽t⩽β
d y d x = ψ ′ ( t ) φ ′ ( t ) \frac{dy}{dx}=\frac{\psi'(t)}{\varphi'(t)} dxdy=φ′(t)ψ′(t)，求 d 2 y d x 2 \frac{d^2y}{dx^2} dx2d2y
{ x = φ ( t ) d y d x = ψ ′ ( t ) φ ′ ( t ) \left\{\begin{array}{l} x=\varphi(t) \\ \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\psi^{\prime}(t)}{\varphi^{\prime}(t)} \end{array}\right. {x=φ(t) dxdy=φ′(t)ψ′(t)
d 2 y d x 2 = d d x ( d y d x ) = ψ ′ ′ ( t ) φ ′ ( t ) − ψ ′ ( t ) φ ′ ′ ( t ) [ φ ′ ( t ) ] 3 \frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{\psi^{\prime \prime}(t) \varphi^{\prime}(t)-\psi^{\prime}(t) \varphi^{\prime \prime}(t)}{\left[\varphi^{\prime}(t)\right]^{3}} dx2d2y=dxd(dxdy)=[φ′(t)]3ψ′′(t)φ′(t)−ψ′(t)φ′′(t)
【例4.5.8】【旋轮线（摆线）】 { x = t − sin t , y = 1 − cos t , 0 ⩽ t ⩽ 2 π \left\{\begin{array}{l} x=t-\sin t, \\ y=1-\cos t, \end{array} \quad 0 \leqslant t \leqslant 2 \pi\right. {x=t−sint,y=1−cost,0⩽t⩽2π， d y d x = sin t 1 − cos t = cot t 2 \frac{dy}{dx}=\frac{\sin t}{1-\cos t}=\cot \frac{t}{2} dxdy=1−costsint=cot2t，求 d 2 y d x 2 ∣ t = π \frac{d^2y}{dx^2}|_{t=\pi} dx2d2y∣t=π.
【解】 d 2 y d x 2 = ( cot t 2 ) ′ 1 − cos t = − 1 2 csc 2 t 2 2 sin 2 t 2 = − 1 4 csc 2 t 2 \frac{d^2y}{dx^2}=\frac{(\cot \frac{t}{2})'}{1-\cos t}=\frac{-\frac{1}{2}\csc^2 \frac{t}{2}}{2\sin^2\frac{t}{2}}=-\frac{1}{4}\csc^2\frac{t}{2} dx2d2y=1−cost(cot2t)′=2sin22t−21csc22t=−41csc22t
所以 d 2 y d x 2 ∣ t = π = − 1 4 ⋅ 1 = − 1 4 \frac{d^2y}{dx^2}|_{t=\pi}=-\frac{1}{4}\cdot1=-\frac{1}{4} dx2d2y∣t=π=−41⋅1=−41
4.5.2 高阶微分
y = f ( x ) , d y = f ′ ( x ) d x ( Δ y = f ′ ( x ) Δ x + o ( Δ x ) ) y=f(x),dy=f'(x)dx(\Delta y=f'(x)\Delta x+o(\Delta x)) y=f(x),dy=f′(x)dx(Δy=f′(x)Δx+o(Δx))，若 f ′ ( x ) ≠ 0 ， Δ y ∼ f ′ ( x ) Δ x f'(x)\ne 0，\Delta y\sim f'(x)\Delta x f′(x)=0，Δy∼f′(x)Δx
d x dx dx是跟 x x x无关的，对 x x x的函数求微分， d x dx dx相对于 x x x来说是“常数”
, d 2 y = d ( d y ) = d ( f ′ ( x ) d x ) = d ( f ′ ( x ) ) d x = f ′ ′ ( x ) d x ⋅ d x = f ′ ′ ( x ) d x 2 ,d^2y = d(dy)=d(f'(x)dx)=d(f'(x))dx=f''(x)dx\cdot dx=f''(x)dx^2 ,d2y=d(dy)=d(f′(x)dx)=d(f′(x))dx=f′′(x)dx⋅dx=f′′(x)dx2
d 3 y = d ( d 2 y ) = d ( f ′ ′ ( x ) d x 2 ) = f ′ ′ ′ ( x ) d x 3 d^3y=d(d^2y)=d(f''(x)dx^2)=f'''(x)dx^3 d3y=d(d2y)=d(f′′(x)dx2)=f′′′(x)dx3
…
d n y = d ( d n − 1 y ) = d ( f ( n − 1 ) ( x ) d x n − 1 ) = f ( n ) ( x ) d x n d^ny=d(d^{n-1}y)=d(f^{(n-1)}(x)dx^{n-1})=f^{(n)}(x)dx^n dny=d(dn−1y)=d(f(n−1)(x)dxn−1)=f(n)(x)dxn
f ( n ) ( x ) = d n y d x n f^{(n)}(x)=\frac{d^ny}{dx^n} f(n)(x)=dxndny
高阶微分没有形式不变性
y = f ( u ) , u = g ( x ) , d y = f ′ ( u ) d u y=f(u),u=g(x),dy=f'(u)du y=f(u),u=g(x),dy=f′(u)du
若 y = f ( u ) , u y=f(u),u y=f(u),u是自变量， d y = f ′ ( u ) d u , d 2 y = f ′ ′ ( u ) d u 2 dy=f'(u)du,d^2y=f''(u)du^2 dy=f′(u)du,d2y=f′′(u)du2
d 2 y = [ f ( g ( x ) ) ] ′ ′ d x 2 = ( f ′ ( u ) g ′ ( x ) ) ′ d x 2 = ( f ′ ′ ( u ) [ g ′ ( x ) ] 2 + f ′ ( u ) g ′ ′ ( x ) ) d x 2 = f ′ ′ ( u ) d u 2 + f ′ ( u ) d 2 u ≠ f ′ ′ ( u ) d u 2 d^2y=[f(g(x))]''dx^2=(f'(u)g'(x))'dx^2=(f''(u)[g'(x)]^2+f'(u)g''(x))dx^2=f''(u)du^2+f'(u)d^2u\ne f''(u)du^2 d2y=[f(g(x))]′′dx2=(f′(u)g′(x))′dx2=(f′′(u)[g′(x)]2+f′(u)g′′(x))dx2=f′′(u)du2+f′(u)d2u=f′′(u)du2
所以高阶微分没有形式不变性
若 u = x u=x u=x， d 2 u = 0 d^2u=0 d2u=0
【例4.5.9】求 y = e sin x y=e^{\sin x} y=esinx的二阶微分。
【解】解法1： d y = ( e sin x ) ′ ′ d x 2 = ( e sin x cos x ) ′ d x 2 = ( e sin x cos 2 x − e sin x sin x ) d x 2 = e sin x ( cos 2 x − sin x ) d x 2 d^y=(e^{\sin x})''dx^2=(e^{\sin x}\cos x)'dx^2=(e^{\sin x}\cos^2 x-e^{\sin x}\sin x)dx^2=e^{\sin x}(\cos^2 x- \sin x)dx^2 dy=(esinx)′′dx2=(esinxcosx)′dx2=(esinxcos2x−esinxsinx)dx2=esinx(cos2x−sinx)dx2
解法2： d y = f ′ ′ ( u ) d u 2 + f ′ ( u ) d 2 u , u = sin x , f ( u ) = e u , d u = cos x d x , d 2 u = − sin x d x 2 d^y=f''(u)du^2+f'(u)d^2u,u=\sin x,f(u)=e^u,du=\cos xdx,d^2u=-\sin xdx^2 dy=f′′(u)du2+f′(u)d2u,u=sinx,f(u)=eu,du=cosxdx,d2u=−sinxdx2
d 2 y = e sin x cos 2 x d x 2 + e sin x ( − sin x ) d x 2 = e sin x ( cos 2 x − sin x ) d x 2 d^2y=e^{\sin x}\cos^2 xdx^2+e^{\sin x}(-\sin x)dx^2=e^{\sin x}(\cos^2 x- \sin x)dx^2 d2y=esinxcos2xdx2+esinx(−sinx)dx2=esinx(cos2x−sinx)dx2
