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Next: Chain Rule and the Up: 多變數函數的微分學 Previous: Limit and Continuity

偏導函數

在本節中我們討論多變數函數的導函數. 單變數函數的導數可以解釋微函數的變化率. 多變數函數的變化情形會依方向的不同而有差異. 所以討論多變數函數在一點 p 的導函數時, 需要就過 p 的各個曲線考慮. 實際做法如下: 設 D $\mathbb{R} ^{n} $ 中的一區域, f 為定義在 D 上的函數, $p \in D$,

\begin{displaymath}C: x= \gamma (t), \,\,\,\,\, t \in [a,b]
\end{displaymath}

為通過 p 的一曲線, $p= \gamma (t_{0} )$. 若 $f( \gamma (t)) $可微, 則

\begin{displaymath}\left. \frac{d}{dt} f( \gamma (t)) \right\vert _{t=t_{0} }
\end{displaymath}

稱為 f 沿 Cp 處的函數.

u $\mathbb{R} ^{n} $ 中的向量, $ \delta >0 $. 假定對一切 $t \in \mathbb{R} $, $\vert t\vert \leq \delta $, 均有 $p+tu \in D$. 考慮線段 $C: x=p+tu,\, t \in [ \delta , \delta ]$. 則 f 沿 Cp 處的導函數稱為 fp 處對 u 的導函數, 以下以 Duf(p) 表示. 換言之,

\begin{displaymath}D_{u}f(p)=\lim_{t \rightarrow 0} \frac{f(p+tu)-f(p)}{t}.
\end{displaymath}

假若 $u \neq 0$, 則 $D_{u/ \Vert u \Vert }f(p)$ 稱為 fp 沿 u 方向的方向導函數 (directional derivative). 因為 Dauf(p)=aDuf(p) 對任意實數 a 均成立, 所以若能知道一函數的諸方向導函數, 則它關於任意向量的導函數便都清楚了.

u 為沿第 k 個座標軸的單位向量, $(1 \leq k \leq n)$, 即 $u=(0,0,\ldots,0,1,$ $0,\ldots,0)$, 式中的 1 在第 k 個位置, 我們便把 Duf(p) 表成

\begin{displaymath}\frac{ \partial f}{ \partial x_{k} } (p) \hbox to 1truecm{\hf...
...1truecm{\hfill} \textup{或} \hbox to 1truecm{\hfill} f_{k} (p)
\end{displaymath}

之形, 而稱之為 partial derivative (偏導函數) of f with respect to xk at p.

例 4   If $\displaystyle f(x,y)=2x^{2} -x^{3} y^{4}$, calculate $\displaystyle\frac{\partial f}{\partial x},\,\frac{\partial f}{\partial y}$and Duf(x,y) for u=(a,b).

解. 計算 $ \displaystyle\frac{\partial f}{\partial x} $ 時, 視 y 為參數, 並保持不變. 乃有

\begin{displaymath}\frac{\partial f}{\partial x} =4x-3x^{2} y^{4}.
\end{displaymath}

仿此 $\displaystyle\frac{\partial f}{\partial y} =-4x^{3} y^{3}$. 對 u=(a,b), 令 (x,y) 為任意定點. 設

g(t)=f(x+ta, y+tb)=2(x+ta)2-(x+ta)3 (y+tb)4.

Duf(x,y)=g'(0)=4ax-3ax2 y4 -4bx3 y3.

例 5   Let f(x,y)=(x2 +y2 )e-x2 -y2. Find $ \displaystyle\frac{\partial f}{\partial x} , \frac{\partial f}{\partial y}$and D(a, b) f(x, y).

Solution. Let u=(a,b) . To calculae D(a, b)f(x, y) , we put

g(t)=f((x,y)+tu)=[(x+ta)2 +(y+tb)2 ]e-(x+ta)2 -(y+tb)2

Then Du f(x, y)=g'(0)=2(ax+by)e-x2 -y2 (1-x2 -y2 ). In particular, if u=(a,b)=(1,0), we have

\begin{displaymath}\frac{\partial f}{\partial x} (x, y)=2ye^{-x^{2} -y^{2} }(1-x^{2} -y^{2} ),
\end{displaymath}

and if u=(a,b)=(0,1), we have

\begin{displaymath}\frac{\partial f}{\partial x} (x, y)=2ye^{-x^{2} -y^{2} }(1-x^{2} -y^{2} ).
\end{displaymath}

例 6   Let

\begin{displaymath}f(x,y) = \left\{ \begin{array}{ll}
\displaystyle\frac{y^{3} }...
...smallskip }
0, & \textup{if }(x,y)=(0,0).
\end{array}\right.
\end{displaymath}

Compute $\displaystyle\frac{\partial f}{\partial x}(0,0),\frac{\partial f}{\partial y}(0, 0)$and $D_{(\cos \theta,\sin\ theta)}f(0,0) $.

Solution. Since f(x,0) = 0 and f(0,y) = y, we have $\displaystyle\frac{\partial f}{\partial x} (0,0)= 0 $ and $\displaystyle\frac{\partial f}{\partial y} (0, 0)=1 $. Now $f(t\cos\theta,t\sin\theta)=t\sin^{3} \theta $. Thus

\begin{displaymath}D_{\cos\theta ,\sin\theta}f(0,0) = \sin^{3}\theta.
\end{displaymath}

Looking back to these examples, we see immediately that for the functions fin Ex. 4 and Ex. 5, we have

\begin{displaymath}D_{(a,b)} f(0,0) = aD_{(1,0)} f + bD_{(0,1)} f = a\frac{\partial{f} }{\partial{x} } + b\frac{\partial{f} }{\partial{y} } ,
\end{displaymath}

but for the function f in Ex. 6,

\begin{displaymath}D_{(a,b)} f(0,0)\neq a\frac{\partial{f} }{\partial{x} } +b\frac{\partial{f} }{\partial{y} }
\end{displaymath}

if $a\neq 0$ . This problem will be studied in the next section.


next up previous
Next: Chain Rule and the Up: 多變數函數的微分學 Previous: Limit and Continuity

1999-06-28