课堂笔记-多元函数微分

多元函数微分

1.基本概念

一，多元函数的定义

二，多元函数的极限与连续

1.lim⁡P→P0f(P)=A⇔∀ε>0,when:0<∣P−P0∣<δ,st:∣f(P)−A∣<ε1. \lim _{P \rightarrow P_{0}} f(P)=A \Leftrightarrow \forall \varepsilon>0,when:0<\left|P-P_{0}\right|<\delta,st:|f(P)-A|<\varepsilon 1. P→P0limf(P)=A ⇔ ∀ε>0,when:0<∣P−P0∣<δ,st:∣f(P)−A∣<ε

注意P→P0方式是任意的，如果任意性不满足，则极限不存在注意 P \rightarrow P_{0} 方式是任意的，如果任意性不满足，则极限不存在注意P→P0方式是任意的，如果任意性不满足，则极限不存在

连续lim⁡P→P0f(P)=f(P0)注意：闭区域上连续函数有与闭区间上连续函数类似的性质.连续 \lim _{P \rightarrow P_{0}} f(P)=f\left(P_{0}\right) 注意：闭区域上连续函数有与闭区间上连续函数类似的性质. 连续P→P0limf(P)=f(P0)注意：闭区域上连续函数有与闭区间上连续函数类似的性质.

三，偏导数的定义

一个多变量的函数的偏导数，就是它关于其中一个变量的导数而保持其他变量恒定
f(x,y)在点(x0,y0)处对x的偏导数:lim⁡Δx→0f(x0+Δx,y0)−f(x0,y0)Δx（极限存在时）f(x,y)在点(x_{0},y_{0})处对x的偏导数:\lim _{\Delta x \rightarrow 0} \frac{f\left(x_{0}+\Delta x, y_{0}\right)-f\left(x_{0}, y_{0}\right)}{\Delta x}（极限存在时） f(x,y)在点(x0,y0)处对x的偏导数:Δx→0limΔxf(x0+Δx,y0)−f(x0,y0)（极限存在时）

偏导数存在⇎连续或极限存在偏导数存在\nLeftrightarrow连续或极限存在偏导数存在⇎连续或极限存在

四，全微分

Δz=AΔx+BΔy+o(ρ)⟹dz=∂z∂xdx+∂z∂ydy\Delta z=A \Delta x+B \Delta y+o(\rho) \Longrightarrow d z=\frac{\partial z}{\partial x} d x+\frac{\partial z}{\partial y} d y Δz=AΔx+BΔy+o(ρ)⟹dz=∂x∂zdx+∂y∂zdy

函数连续⇎偏导存在.函数连续 \nLeftrightarrow 偏导存在. 函数连续⇎偏导存在.

函数可微⇒偏导存在(一阶,二阶).偏导存在⇏函数可微函数可微 \Rightarrow 偏导存在(一阶,二阶). 偏导存在 \nRightarrow 函数可微函数可微⇒偏导存在(一阶,二阶).偏导存在⇏函数可微

函数可微⇒函数连续.函数连续⇏函数可微.函数可微 \Rightarrow 函数连续. 函数连续 \nRightarrow 函数可微. 函数可微⇒函数连续.函数连续⇏函数可微.

1.如果∂z∂x，∂z∂y连续⇒可微.1.如果\frac{\partial z}{\partial x}，\frac{\partial z}{\partial y}连续 \Rightarrow可微. 1.如果∂x∂z，∂y∂z连续⇒可微.

2.若fxy与fyx在(x0,y0)的某邻域内连续⇒fxy=fyx.2.若f_{x y}与f_{y x}在(x_{0},y_{0})的某邻域内连续\Rightarrow f_{x y}=f_{y x}. 2.若fxy与fyx在(x0,y0)的某邻域内连续⇒fxy=fyx.

五，习题

1.f(x,y)={xyx2−y2x2+y2,x2+y2≠00,x2+y2=0,proof:fxy(0,0)≠fyx(0,0)1.\quad \quad f(x, y)=\left\{\begin{array}{ll}{x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}},} & {x^{2}+y^{2} \neq 0} \\ {0,} & {x^{2}+y^{2}=0}\end{array}\right. ,proof:f_{xy}(0,0)\neq f_{yx}(0,0) 1.f(x,y)={xyx2+y2x2−y2,0,x2+y2̸=0x2+y2=0,proof:fxy(0,0)̸=fyx(0,0)

x=0,∂f(0,y)∂x=lim⁡Δx→0f(Δx,y)−f(0,y)Δx=lim⁡Δx→0Δxy(Δx)2−y2(Δx)2+y2−0Δx=−yx=0, \quad \frac{\partial f(0, y)}{\partial x}=\lim _{\Delta x \rightarrow 0} \frac{f(\Delta x , y)-f(0, y)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x y \frac{(\Delta x)^{2}-y^{2}}{(\Delta x)^{2}+y^{2}}-0}{\Delta x}=-y x=0,∂x∂f(0,y)=Δx→0limΔxf(Δx,y)−f(0,y)=Δx→0limΔxΔxy(Δx)2+y2(Δx)2−y2−0=−y

fxy(0,0)=∂∂y(∂f∂x)=∂2f(0,0)∂x∂y=lim⁡y→0fx(0,Δy)−fx(0,0)Δy=−1f_{x y}(0,0)=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial^{2} f(0,0)}{\partial x \partial y}=\lim _{y \rightarrow 0} \frac{f_{x}(0,\Delta y)-f_{x}(0,0)}{\Delta y}=-1 fxy(0,0)=∂y∂(∂x∂f)=∂x∂y∂2f(0,0)=y→0limΔyfx(0,Δy)−fx(0,0)=−1

y=0,fy(x,0)=∂f(x,0)∂y=lim⁡Δy→0f(x,Δy)−f(x,0)Δy=lim⁡Δy→0xΔy(x2−(Δy)2)x2+(Δy)2−0Δy=xy=0, \quad f_{y}(x, 0)= \frac{\partial f(x, 0)}{\partial y}=\lim _{\Delta y \rightarrow 0} \frac{f(x, \Delta y)-f(x, 0)}{\Delta y}=\lim _{\Delta y \rightarrow 0} \frac{\frac{x \Delta y\left(x^{2}-(\Delta y)^{2}\right)}{x^{2}+(\Delta y)^{2}}-0}{\Delta y}=x y=0,fy(x,0)=∂y∂f(x,0)=Δy→0limΔyf(x,Δy)−f(x,0)=Δy→0limΔyx2+(Δy)2xΔy(x2−(Δy)2)−0=x

fyx(0,0)=∂2f(0,0)∂y∂x=lim⁡Δx→0fy(Δx,0)−fy(0,0)Δx=lim⁡Δx→0Δx−0Δx=1f_{y x}(0,0)=\frac{\partial^{2} f(0,0)}{\partial y \partial x}=\lim _{\Delta x \rightarrow 0} \frac{f_{y}(\Delta x, 0)-f_{y}(0,0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x-0}{\Delta x}=1 fyx(0,0)=∂y∂x∂2f(0,0)=Δx→0limΔxfy(Δx,0)−fy(0,0)=Δx→0limΔxΔx−0=1

2.prooff(x,y)={x2yx4+y2,x4+y2≠00,x4+y2=0在(0,0)的极限不存在.2.proof \quad \quad f(x, y)=\left\{\begin{array}{ll}{\frac{x^{2} y}{x^{4}+y^{2}},} & {x^{4}+y^{2} \neq 0} \\ {0,} & {x^{4}+y^{2}=0}\end{array}\right.在(0,0)的极限不存在. 2.prooff(x,y)={x4+y2x2y,0,x4+y2̸=0x4+y2=0在(0,0)的极限不存在.
设y=kx2,lim⁡(x,y)→(0,0)f(x,y)=lim⁡(x,y)→(0,0)kx4x4+k2x4=k1+k2设y=kx^2,\quad\quad \lim _{(x, y) \rightarrow(0,0)} f(x, y)=\lim _{(x, y) \rightarrow(0,0)} \frac{k x^{4}}{x^{4}+k^{2} x^{4}}=\frac{k}{1+k^{2}} 设y=kx2,(x,y)→(0,0)limf(x,y)=(x,y)→(0,0)limx4+k2x4kx4=1+k2k

3.prooff(x,y)={x2y1+αx4+y2,(x,y)≠(0,0)0,(x,y)=(0,0)在(0,0)处连续，其中α>03.proof\quad \quad f(x, y)=\left\{\begin{array}{ll}{\frac{x^{2} y^{1+\alpha}}{x^{4}+y^{2}},} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right.在(0,0)处连续，其中\alpha>0 3.prooff(x,y)={x4+y2x2y1+α,0,(x,y)̸=(0,0)(x,y)=(0,0)在(0,0)处连续，其中α>0
∣f(x,y)−f(0,0)∣=x2∣y∣1+αx4+y2≤x2∣y∣1+α2x2∣y∣=∣y∣α→0tip:2x2∣y∣≤x4+y2|f(x, y)-f(0,0)|=\frac{x^{2}|y|^{1+\alpha }} {x^{4}+y^{2}} \leq \frac{x^{2}|y|^{1+\alpha}}{2 x^{2}|y|}=|y|^\alpha\rightarrow 0 \quad\quad tip:2x^2|y|\leq x^4+y^2 ∣f(x,y)−f(0,0)∣=x4+y2x2∣y∣1+α≤2x2∣y∣x2∣y∣1+α=∣y∣α→0tip:2x2∣y∣≤x4+y2

4.求u=xy2z3沿着曲线x=t,y=2t2,z=−2t4在点M(1,2,−2)处切线方向的方向导数.4.\quad \quad 求u=x y^{2} z^{3}沿着曲线x=t,y=2t^2,z=-2t^4在点M(1,2,-2)处切线方向的方向导数. 4.求u=xy2z3沿着曲线x=t,y=2t2,z=−2t4在点M(1,2,−2)处切线方向的方向导数.

tip:梯度grad⁡u={∂u∂x,∂u∂v,∂u∂z}tip:梯度 \operatorname{grad} u=\left\{\frac{\partial u}{\partial x}, \frac{\partial u}{\partial v}, \frac{\partial u}{\partial z}\right\} tip:梯度gradu={∂x∂u,∂v∂u,∂z∂u}

∂u∂l⃗∣P0=∂u∂x∣p0cos⁡α+∂u∂y∣p0cos⁡β+∂u∂z∣p0cos⁡γ\left.\frac{\partial u}{\partial \vec{l}}\right|_{P_{0}} = \left.\frac{\partial u}{\partial x} \right |_{p_{0}} \cos \alpha + \left.\frac{\partial u}{\partial y} \right |_{p_{0}} \cos \beta+\left.\frac{\partial u}{\partial z} \right |_{p_{0}} \cos \gamma ∂l∂u∣∣∣∣P0=∂x∂u∣∣∣∣p0cosα+∂y∂u∣∣∣∣p0cosβ+∂z∂u∣∣∣∣p0cosγ