Now we consider \(f:\mathbb{R}^N\to\mathbb{R}\)
Definition 37 If the limit
| \[\lim_{h\to0}\frac{f(x_1,x_2,\ldots,x_{k}+h,\ldots,x_N)-f(x_1,x_2,\ldots,x_k,\ldots,x_N)}{h}\] |
exists, then it is called the partial derivative
| \[\frac{\partial f}{\partial x_k}\] |
of \(f\) with respect to \(x_k\).
In vector
notation: \begin{eqnarray*} \vec{x}&=&\vect{x_1\\\vdots\\x_N}\qquad\vec{e}_k=\vect{0\\0\\\vdots\\0\\1\\0\\\vdots\\0}
\mbox{the 1 is at the $k$-th position and } \vec{e_k}\in\mathbb{R}^N\\ \rightarrow\;\frac{\partial f}{\partial x_k}&=&\lim_{h\to0}\frac{f(\vec{x}+h\vec{e}_k)-f(\vec{x})}{h}\;\;\mbox{(note:
$\frac{df}{dx}$ in the case of one variable)}\\ &&\mbox{note that we have now round ''$\partial$'s ''}\end{eqnarray*}
Example: \begin{eqnarray*}F(x_1,x_2)&=&x_1^2+x_2^3\\ \frac{\partial f}{\partial x_1}&=&\lim_{h\to0}\frac{(x_1+h)^2+x^3_2-(x_1^2+x_2^3)}{h}=\lim_{h\to0}\frac{(x_1+h)^2-x_1^2}{h}\\ &=&\lim_{h\to0}\frac{2x_1h+h^2}{h}=\lim_{h\to0}(2x_1+h)=2x_1\\ \mbox{similar: }\;\frac{\partial f}{\partial x_2}&=&3x_2^2\end{eqnarray*}
In general: \begin{eqnarray*}\mbox{partial derivative } \frac{\partial f}{\partial x_k}&\hat=&\mbox{''normal'' derivative with respect to $x_k$ where the other}\\&&\mbox{variables $x_1,\ldots,x_{k-1},x_{k+1},\ldots,x_N$ are kept constant as parameters!}\\ &\Rightarrow&\mbox{partial derivatives totally equal to normal derivatives!}\end{eqnarray*}
Example: \begin{eqnarray*}f(x,y)&=&e^{x^2}y^3\\ \rightarrow\,\frac{\partial f}{\partial x}&=&2xe^{x^2}y^3\\ \frac{\partial f}{\partial y}&=&3y^2e^{x^2}\\ \\ f(x_1,\ldots,x_N)&=&\sqrt{x_1^2+x_2^2+\ldots+x_N^2}=\sqrt{(\vec{r})^2},\;\;\vec{r}=\vect{x_1\\x_2\\\vdots\\x_N}\\ \rightarrow\,\frac{\partial f}{\partial x_k}&=&2x_k\frac{1}{2}(x_1^2+\ldots+x_N^2)^{-\frac{1}{2}}=\frac{x_k}{|\vec{r}|}\end{eqnarray*}
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© J. Carstensen (Math for MS)