편도함수

As $z(x,y)=f(x^2+y^2)$

$$\begin{aligned} \frac{\partial z}{\partial x}(x,y)&=2xf'(x^2+y^2) \\ \frac{\partial z}{\partial y}(x,y)&=2yf'(x^2+y^2) \end{aligned}$$

by the (ordinary single variable) chain rule. So

$$y\frac{\partial z}{\partial x}-x\frac{\partial z}{\partial y} =y(2x)f'(x^2+y^2)-x(2y)f'(x^2+y^2)=0$$

and the differential equation is always satisfied, assuming that $f$ is differentiable, so that the chain rule applies.

By definition

$$\begin{aligned} \frac{\partial f}{\partial x}(0,0) &=\lim_{\Delta x\rightarrow 0}\frac{f(\Delta x,0)-f(0,0)}{\Delta x} \\ &=\lim_{\Delta x\rightarrow 0}\frac{\frac{(\Delta x+2\times 0)^2}{\Delta x+0}-0}{\Delta x} \\ &=\lim_{\Delta x\rightarrow 0}\frac{\Delta x}{\Delta x} \\ &=1 \end{aligned}$$

and

$$\begin{aligned} \frac{\partial f}{\partial y}(0,0) &=\lim_{\Delta y\rightarrow 0}\frac{f(0,\Delta y)-f(0,0)}{\Delta y} \\ &=\lim_{\Delta y\rightarrow 0}\frac{\frac{(0+2\Delta y)^2}{0+\Delta y}-0}{\Delta y} \\ &=\lim_{\Delta y\rightarrow 0}\frac{4\Delta y}{\Delta y} \\ &=4 \end{aligned}$$

(b) $f(x,y)$ is not continuous at $(0,0)$, even though both partial derivatives exist there. To see this, make a change of coordinates from $(x,y)$ to $(X,y)$ with $X=x+y$ (the denominator). Of course, $(x,y)\rightarrow (0,0)$ if and only if $(X,y)\rightarrow (0,0)$. Now watch what happens when $(X,y)\rightarrow(0,0)$ with $X$ a lot smaller than $y$. For example, $X=ay^2$. Then

$$\begin{aligned} \frac{(x+2y)^2}{x+y}=\frac{(X+y)^2}{X}=\frac{(ay^2+y)^2}{ay^2} =\frac{(1+ay)^2}{a}\rightarrow\frac{1}{a} \end{aligned}$$

This depends on $a$. So approaching $(0,0)$ along different paths gives different limits. (You can see the same effect without changing coordinates by sending $(x,y)\rightarrow (0,0)$ with $x=-y+ay^2$.) Even more dramatically, watch what happens when $(X,y)\rightarrow(0,0)$ with $X=y^3$. Then

$$\frac{(x+2y)^2}{x+y}=\frac{(X+y)^2}{X}=\frac{(y^3+y)^2}{y^3} =\frac{{(1+y^2)}^2}{y}\rightarrow\pm\infty$$

Solution 1

Let's start by finding an equation for this surface. Every level curve is a horizontal circle of radius one, so the equation should be of the form

$$(x-f_1)^2+(y-f_2)^2=1$$

where $f_1$ and $f_2$ are functions depending only on $z$. Since the centre of the circle at height $z$ is at position $x=0$, $y=z$, we see that the equation of our surface is

$$x^2+(y-z)^2=1$$

The height of the surface at the point $(x,y)$ is the $z(x,y)$ found by solving that equation. That is,

$$x^2+\big(y-z(x,y)\big)^2=1 \tag{$*$}$$

We differentiate this equation implicitly to find $z_x(x,y)$ and $z_y(x,y)$ at the desired point $(x,y)= (0,-1)$. First, differentiating $(*)$ with respect to $y$ gives

$$\begin{aligned} 0+2\big(y-z(x,y)\big)\big(1- z_y(x,y)\big)&=0 \\ 2(-1-0)\big(1-z_y(0,-1)\big)&=0& & \text{ at } (0,-1,0) \end{aligned}$$

so that the slope looking in the positive $y$ direction is $z_y(0,-1)=1$. Similarly, differentiating $(*)$ with respect to $x$ gives

$$\begin{aligned} 2x+2\big(y-z(x,y)\big)\cdot\big(0-z_x(x,y)\big)&=0 \\ 2x&=2\big(y-z(x,y)\big)\cdot z_x(x,y)\\ z_x(x,y)&=\frac{x}{y-z(x,y)}\\ z_x(0,-1)&=0 &\text{ at } (0,-1,0) \end{aligned}$$

The slope looking in the positive $x$ direction is $z_x(0,-1)=0$.

Solution 2

Standing at $(0,-1,0)$ and looking in the positive $y$ direction, the surface follows the straight line that

• passes through the point $(0,-1,0)$, and
• is parallel to the central line $z=y, x=0$ of the cylinder.

Shifting the central line one unit in the $y$-direction, we get the line $z=y+1$, $x=0$. (As a check, notice that $(0,-1,0)$ is indeed on $z=y+1$, $x=0$.) The slope of this line is 1.

Standing at $(0,-1,0)$ and looking in the positive $x$ direction, the surface follows the circle $x^2+y^2=1$, $z=0$, which is the intersection of the cylinder with the $xy$-plane. As we move along that circle our $z$ coordinate stays fixed at $0$. So the slope in that direction is 0.

No such $f(x,y)$ exists, because if it were to exist, then we would have that $f_{xy}(x,y)=f_{yx}(x,y)$. But

$$\begin{aligned} f_{xy}(x,y)&=\frac{\partial }{\partial y}f_x(x,y)=\frac{\partial }{\partial y}e^y=e^y \\ f_{yx}(x,y)&=\frac{\partial }{\partial x}f_y(x,y)=\frac{\partial }{\partial x}e^x=e^x \end{aligned}$$

are not equal.

(a) We have

$$\begin{aligned} f_x(x,y) &= 2xy^3 & f_{xx}(x,y) &= 2y^3 \\ & & f_{xy}(x,y) &= 6xy^2 & f_{yxy}(x,y) = f_{xyy}(x,y) &= 12xy \end{aligned}$$

(b) We have

$$\begin{aligned} f_x(x,y) &= y^2e^{xy^2} & f_{xx}(x,y) &= y^4e^{xy^2} & f_{xxy}(x,y) &= 4y^3e^{xy^2} + 2xy^5e^{xy^2} \\ & & f_{xy}(x,y) &= 2ye^{xy^2}+2xy^3e^{xy^2} & f_{xyy}(x,y) &= \big(2+4xy^2+6xy^2+4x^2y^4\big)e^{xy^2}\\ & & & & &= \big(2+10xy^2+4x^2y^4\big)e^{xy^2} \end{aligned}$$

(c) We have

$$\begin{aligned} \frac{\partial f}{\partial u}(u,v,w) &= -\frac{1}{(u+2v+3w)^2} \\ \frac{\partial^2 f}{\partial u\,\partial v}(u,v,w) &= \frac{4}{(u+2v+3w)^3} \\ \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(u,v,w) &= -\frac{36}{(u+2v+3w)^4} \end{aligned}$$

In particular

$$\begin{aligned} \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(3,2,1) &= -\frac{36}{(3+2\times 2+3\times 1)^4} = -\frac{36}{10^4} = -\frac{9}{2500} \end{aligned}$$

Let $f(x,y)=\sqrt{x^2+5y^2}$. Then

$$\begin{aligned} f_x&=\frac{x}{\sqrt{x^2+5y^2}} & f_{xx}&=\frac{1}{\sqrt{x^2+5y^2}}-\frac{1}{2}\frac{(x)(2x)}{(x^2+5y^2)^{3/2}} & f_{xy}&=-\frac{1}{2}\frac{(x)(10y)}{(x^2+5y^2)^{3/2}} \\ f_y&=\frac{5y}{\sqrt{x^2+5y^2}} & f_{yy}&=\frac{5}{\sqrt{x^2+5y^2}}-\frac{1}{2}\frac{(5y)(10y)}{(x^2+5y^2)^{3/2}}& f_{yx}&=-\frac{1}{2}\frac{(5y)(2x)}{(x^2+5y^2)^{3/2}} \end{aligned}$$

Simplifying, and in particular using that $\frac{1}{\sqrt{x^2+5y^2}} =\frac{x^2+5y^2}{(x^2+5y^2)^{3/2}}$,

$$f_{xx}=\frac{5y^2}{(x^2+5y^2)^{3/2}}\qquad f_{xy}=f_{yx}=-\frac{5xy}{(x^2+5y^2)^{3/2}}\qquad f_{yy}=\frac{5x^2}{(x^2+5y^2)^{3/2}}$$

(a) As $f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ is independent of $z$, we have $f_z(x,y,z) = 0$ and hence

$$f_{xyz}(x,y,z) =f_{zxy}(x,y,z) =0$$

(b) Write $u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$, $v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ and $w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)$. Then

• As $u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ is independent of $z$, we have $u_z(x,y,z) = 0$ and hence $u_{xyz}(x,y,z) =u_{zxy}(x,y,z) =0$
• As $v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ is independent of $y$, we have $v_y(x,y,z) = 0$ and hence $v_{xyz}(x,y,z) =v_{yxz}(x,y,z) =0$
• As $w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)$ is independent of $x$, we have $w_x(x,y,z) = 0$ and hence $w_{xyz}(x,y,z) =0$