《문제》편도함수

1. 다음 함수의 원점 \((0,\,0)\)에서의 편미분계수를 구하여라.

(1) \(f(x,\,y)=\sqrt{|xy|}\)          (2) \(f(x,\,y)=\sqrt{x^2+y^2}\)          (3) \(f(x,\,y)=\begin{cases}\frac{xy}{\sqrt{x^2+y^2}}&(x,\,y)\ne(0,\,0)\\0&(x,\,y)=(0,\,0)\end{cases}\)

<풀이>

(1) \(f_x(0,\,0)=\lim_{h\to0}\frac{f(h,\,0)-f(0,\,0)}{h}=0,\,f_y(0,\,0)=\lim_{h\to0}\frac{f(0,\,h)-f(0,\,0)}{h}=0\)

(2) \(f_x(0,\,0)=\lim_{n\to\pm0}\frac{f(h,\,0)-f(0,\,0)}{h}=\pm1,\,f_y(0\,0)=\lim_{h\to\pm0}\frac{f(0,\,h)-f(0,\,0)}{h}=\pm1\)

     ∴ 원점에서는 편미분계수가 존재하지 않는다.

(3) \(f_x(0,\,0)=\lim_{h\to0}\frac{f(h,\,0)-f(0,\,0)}{h}=0,\,f_y(0,\,0)=\lim_{h\to0}\frac{f(0,\,h)=f(0,\,0)}{h}=0\)

2. 다음 함수를 편미분하여라.

(1) \(f(x,\,y)=x^2+3xy^2-2x+4y+3\)

(2) \(f(x,\,y)=x^3+3x^2y-y^3\)          (3) \(f(x,\,y)=2x+ye^{-x}\)

(4) \(f(x,\,y)=\sqrt{x^2+xy+y^2}\)         (5) \(f(x,\,y)={\rm Sin}^{-1}{x\over y}\)

(6) \(f(x,\,y)={\rm Tan}^{-1}\frac{x+y}{x-y}\)                (7) \(f(x,\,y)=e^{ax}\cos{by}\)

<풀이>

(1) \(f_x(x,\,y)=2x+3y^2-2,\,f_y(x,\,y)=6xy+4\)

(2) \(f_x(x,\,y)=3x^2+6xy,\,f_y(x,\,y)=3x^2-3y^2\)

(3) \(f_x(x,\,y)=2-ye^{-x},\,f_y(x,\,y)=e^{-x}\)

(4) \(f_x(x,\,y)=\frac{2x+y}{2\sqrt{x^2+xy+y^2}},\,f_y(x,\,y)=\frac{x+2y}{2\sqrt{x^2+xy+y^2}}\)

(5) \(f_x(x,\,y)=\frac{1/y}{\sqrt{1-(x/y)^2}}=\begin{cases}{1\over\sqrt{y^2-x^2}}&(y>0)\\{-1\over\sqrt{y^2-x^2}}&(y<0)\end{cases},\,f_y(x,\,y)=\frac{-x/y^2}{\sqrt{1-(x/y)^2}}=\begin{cases}{-x\over\sqrt{y^2-x^2}}&(y>0)\\{x\over\sqrt{y^2-x^2}}&(y<0)\end{cases}\)

(6) \(f_x(x,\,y)=\frac{1}{1+\left(\frac{x+y}{x-y}\right)^2}\cdot\frac{-2y}{(x-y)^2}=\frac{-y}{x^2+y^2},\,f_y(x,\,y)=\frac{1}{1+\left(\frac{x+y}{x-y}\right)^2}\cdot\frac{2x}{(x-y)^2}=\frac{x}{x^2+y^2}\)

(7) \(f_x(x,\,y)=ae^{ax}\cos{by},\,f_y(x,\,y)=-be^{ax}\sin{by}\)

3. 다음의 각각에 대해서 \(dz/dt\)를 구하여라. (합성함수의 편미분법 참조)

(1) \(z=x^2+y^2,\,x=1-\sin{t},\,y=1-\cos{t}\)

(2) \(z={y\over x},\,x=e^t,\,y=\sin{t}\)          (3) \(z={\rm Tan}^{-1}tx,\,x=e^t\)

<풀이>

(1) \(\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}=2x(1-\cos{t})+2y\sin{t}=2t(1-cos{t})\)

(2) \(\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}=-e^{-t}\sin{t}+e^{-t}\cos{t}=e^{-t}(\cos{t}-\sin{t})\)

(3) \(\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial t}=\frac{te^t}{1+t^2e^{2t}}+\frac{e^t}{1+t^2e^{2t}}=\frac{e^t(1+t)}{1+t^2e^{2t}}\)

4. \(x=r\cos\theta,\,y=r\sin\theta\)라 하면, \(x,\,y\)는 \(r,\,\theta\)의 함수이고, 역으로 \(r,\,\theta\)는 \(x,\,y\)의 함수이다.

(1) \(\frac{\partial x}{\partial r}=\frac{\partial r}{\partial x}\)          (2) \(\frac{\partial x}{\partial\theta}=r^2\frac{\partial\theta}{\partial x}\)          (3) \(\frac{\partial y}{\partial r}=\frac{\partial r}{\partial y}\)          (4) \(\frac{\partial y}{\partial\theta}=r^2\frac{\partial\theta}{\partial y}\)

임을 증명하여라.

<풀이>

(1) \(\frac{\partial x}{\partial r}=\cos\theta,\,\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}=\cos\theta\)

(2) \(\frac{\partial x}{\partial\theta}=-r\sin\theta,\,r^2\frac{\partial\theta}{\partial x}=r^2\left(\frac{-y}{x^2+y^2}\right)=-r\sin\theta\)

(3) \(\frac{\partial y}{\partial r}=\sin\theta,\,\frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}=\sin\theta\)

(4) \(\frac{\partial y}{\partial\theta}=r\cos\theta,\,r^2\frac{\partial\theta}{\partial y}=r^2\left(\frac{x}{x^2+y^2}\right)=r\cos\theta\)

5. \(z=f(u),\,u=x^2-y^2\) 일 때 \(y\frac{\partial z}{\partial x}+x\frac{\partial z}{\partial y}=0\) 임을 보여라.

<풀이> \((e^x+e^y)z=e^{xy}\) 이므로 양변을 \(x\)와 \(y\)로 편미분하면

\(\begin{cases}e^xz+(e^x+e^y)\frac{\partial z}{\partial x}=e^{xy}y\\e^yz+(e^x+e^y)\frac{\partial z}{\partial y}=e^{xy}x\end{cases}\)

이 두식의 양변을 더하면 \(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=z(x+y-1)\) 이다.