다변수 미분 전체 문제

(a) We are told that $z(x,y)$ obeys

$$\begin{aligned} z(x,y)\, y - y + x = \ln\big(xy\,z(x,y)\big) \tag{$*$}\end{aligned}$$

for all $(x,y)$ (near $(-1,-2)$). Differentiating $(*)$ with respect to $x$ gives

$$\begin{aligned} y\,\frac{\partial z}{\partial x}(x,y) + 1 = \frac{1}{x} + \frac{\frac{\partial z}{\partial x}(x,y)}{z(x,y)} \implies \frac{\partial z}{\partial x}(x,y) = \frac{\frac{1}{x}-1}{y-\frac{1}{z(x,y)}} \end{aligned}$$

or, dropping the arguments $(x,y)$ and multiplying both the numerator and denominator by $xz$,

$$\begin{aligned} \frac{\partial z}{\partial x} = \frac{z-xz}{xyz-x} = \frac{z(1-x)}{x(yz-1)} \end{aligned}$$

Differentiating $(*)$ with respect to $y$ gives

$$\begin{aligned} z(x,y)+y\,\frac{\partial z}{\partial y}(x,y) - 1 = \frac{1}{y} + \frac{\frac{\partial z}{\partial y}(x,y)}{z(x,y)} \implies \frac{\partial z}{\partial y}(x,y) = \frac{\frac{1}{y}+1-z(x,y)}{y-\frac{1}{z(x,y)}} \end{aligned}$$

or, dropping the arguments $(x,y)$ and multiplying both the numerator and denominator by $yz$,

$$\begin{aligned} \frac{\partial z}{\partial y} = \frac{z+yz-yz^2}{y^2z-y} = \frac{z(1+y-yz)}{y(yz-1)} \end{aligned}$$

(b) When $(x,y,z) = (-1, -2, 1/2)$,

$$\begin{aligned} \frac{\partial z}{\partial x}(-1,-2) &= \left.\frac{\frac{1}{x}-1}{y-\frac{1}{z}} \right|_{(x,y,z) = (-1, -2, 1/2)} =\frac{\frac{1}{-1}-1}{-2-2} =\frac{1}{2} \\ \frac{\partial z}{\partial y}(-1,-2) &= \left.\frac{\frac{1}{y}+1-z}{y-\frac{1}{z}} \right|_{(x,y,z) = (-1, -2, 1/2)} =\frac{\frac{1}{-2}+1-\frac{1}{2}}{-2-2} =0 \end{aligned}$$

We are told that the four variables $T$, $U$, $V$, $W$ obey the single equation $(TU-V)^2 \ln(W-UV) = \ln 2$. So they are not all independent variables. Roughly speaking, we can treat any three of them as independent variables and solve the given equation for the fourth as a function of the three chosen independent variables.

We are first asked to find $\frac{\partial U}{\partial T}$. This implicitly tells to treat $T$, $V$ and $W$ as independent variables and to view $U$ as a function $U(T,V,W)$ that obeys

$$\big(T\, U(T,V,W)-V\big)^2 \ln\big(W-U(T,V,W)\,V\big) = \ln 2 \tag{E1}$$

for all $(T, U, V, W)$ sufficiently near $(1, 1, 2, 4)$. Differentiating (E1) with respect to $T$ gives

$$\begin{aligned} &2\big(T\, U(T,V,W)-V\big) \left[ U(T,V,W) +T\ \frac{\partial U}{\partial T}(T,V,W)\right] \ln\big(W-U(T,V,W)\,V\big) \\ & -\big(T\, U(T,V,W)-V\big)^2 \frac{1}{W-U(T,V,W)\,V}\frac{\partial U}{\partial T}(T,V,W)\,V = 0 \end{aligned}$$

In particular, for $(T, U, V, W)=(1, 1, 2, 4)$,

$$\begin{aligned} &2\big((1)(1)-2\big) \left[ 1 +(1)\frac{\partial U}{\partial T}(1,2,4)\right] \ln\big(4-(1)(2)\big) \\ & -\big((1)(1)-2\big)^2 \frac{1}{4-(1)(2)}\frac{\partial U}{\partial T}(1,2,4)\,(2) = 0 \end{aligned}$$

This simplifies to

$$\begin{aligned} -2\left[ 1 +\frac{\partial U}{\partial T}(1,2,4)\right] \ln(2) -\frac{\partial U}{\partial T}(1,2,4)=0 \implies \frac{\partial U}{\partial T}(1,2,4) = -\frac{2\ln(2)}{1+2\ln(2)} \end{aligned}$$

We are then asked to find $\frac{\partial T}{\partial V}$. This implicitly tells to treat $U$, $V$ and $W$ as independent variables and to view $T$ as a function $T(U,V,W)$ that obeys

$$\big(T(U,V,W)\, U-V\big)^2 \ln\big(W-U\,V\big) = \ln 2 \tag{E2}$$

for all $(T, U, V, W)$ sufficiently near $(1, 1, 2, 4)$. Differentiating (E2) with respect to $V$ gives

$$\begin{aligned} &2\big(T(U,V,W)\, U-V\big)\ \left[\frac{\partial T}{\partial V}(U,V,W)\ U-1\right] \ln\big(W-U\,V\big) \\ & -\big(T(U,V,W)\, U-V\big)^2 \frac{U}{W-U\,V} = 0 \end{aligned}$$

In particular, for $(T, U, V, W)=(1, 1, 2, 4)$,

$$\begin{aligned} &2\big((1)(1)-2\big) \left[ (1)\frac{\partial T}{\partial V}(1,2,4)-1\right] \ln\big(4-(1)(2)\big) \\ & -\big((1)(1)-2\big)^2 \frac{1}{4-(1)(2)} = 0 \end{aligned}$$

This simplifies to

$$\begin{aligned} -2\left[\frac{\partial T}{\partial V}(1,2,4)-1\right] \ln(2) -\frac{1}{2}=0 \implies \frac{\partial T}{\partial V}(1,2,4) = 1 -\frac{1}{4\ln(2)} \end{aligned}$$

The function

$$\begin{aligned} u(\rho, r,\theta) &= \big[\rho r\cos\theta\big]^2 +\big[\rho r\sin\theta\big] \rho r \\ &=\rho^2 r^2\cos^2\theta +\rho^2 r^2\sin\theta \end{aligned}$$

So

$$\begin{aligned} \frac{\partial u}{\partial r}(\rho, r,\theta) &=2 \rho^2 r\cos^2\theta +2 \rho^2 r\sin\theta \end{aligned}$$

and

$$\begin{aligned} \frac{\partial u}{\partial r}(2, 3,\pi/2) &=2 (2^2) (3) (0)^2 +2 (2^2) (3) (1) =24 \end{aligned}$$

By definition

$$f_x(x_0,y_0)=\lim_{\Delta x\rightarrow 0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x} \qquad f_y(x_0,y_0)=\lim_{\Delta y\rightarrow 0}\frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y}$$

Setting $x_0=y_0=0$,

$$\begin{aligned} f_x(0,0)&=\lim_{\Delta x\rightarrow 0}\frac{f(\Delta x,0)-f(0,0)}{\Delta x} &=\lim_{\Delta x\rightarrow 0}\frac{f(\Delta x,0)}{\Delta x} &=\lim_{\Delta x\rightarrow 0}\frac{((\Delta x)^2-2\times0^2)/(\Delta x-0)}{\Delta x} \\ &=\lim_{\Delta x\rightarrow 0}1 =1\\ f_y(0,0)&=\lim_{\Delta y\rightarrow 0}\frac{f(0,\Delta y)-f(0,0)}{\Delta y} &=\lim_{\Delta y\rightarrow 0}\frac{f(0,\Delta y)}{\Delta y} &=\lim_{\Delta y\rightarrow 0}\frac{(0^2-2(\Delta y)^2)/(0-\Delta y)}{\Delta y} \\ &=\lim_{\Delta y\rightarrow 0}2 =2 \end{aligned}$$

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

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