다변수 미분 전체 문제

If $f_y(0,0)<0$, then $f(0,y)$ decreases as $y$ increases from $0$. Thus moving in the positive $y$ direction takes you downhill. This means you aren't at the lowest point in a valley, since you can still move downhill. On the other hand, as $f_y(0,0)<0$, $f(0,y)$ also decreases as $y$ increases towards $0$ from slightly negative values. Thus if you move in the negative $y$-direction from $y=0$, your height $z$ will increase. So you are not at a locally highest point—you're not at a summit.

(a) By definition

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

(b) By definition

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

(c) By definition

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} f(t,t)\Big|_{t=0} &=\lim_{t\rightarrow 0}\frac{f(t,t)-f(0,0)}{t} \\ &=\lim_{t\rightarrow 0}\frac{\frac{(t^2)(t)}{t^2+t^2}-0}{t} \\ &=\lim_{t\rightarrow 0}\frac{t/2}{t} \\ &=\frac{1}{2} \end{aligned}$$

(a)

$$\begin{aligned} f_x(x,y,z)&=3x^2y^4z^5 & f_x(0,-1,-1)&=0\\ f_y(x,y,z)&=4x^3y^3z^5 & f_y(0,-1,-1)&=0\\ f_z(x,y,z)&=5x^3y^4z^4 & f_z(0,-1,-1)&=0 \end{aligned}$$

(b)

$$\begin{aligned} w_x(x,y,z)&=\frac{yz e^{xyz}}{1+e^{xyz}} & w_x(2,0,-1)&=0\\ w_y(x,y,z)&=\frac{xz e^{xyz}}{1+e^{xyz}} & w_y(2,0,-1)&=-1\\ w_z(x,y,z)&=\frac{xy e^{xyz}}{1+e^{xyz}} & w_z(2,0,-1)&=0 \end{aligned}$$

(c)

$$\begin{aligned} f_x(x,y)&=-\frac{x}{(x^2+y^2)^{3/2}} & f_x(-3,4)&=\frac{3}{125}\\ f_y(x,y)&=-\frac{y}{(x^2+y^2)^{3/2}} & f_y(-3,4)&=-\frac{4}{125} \end{aligned}$$

(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 $x_0=y_0=0$,

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