数学题在线解答

问题:设f(x,y) = ysin(1/x^2+y^2), 考察函数f在原点(0,0)的偏导数

解答

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

考察函数f在原点(0,0)的偏导数

二元函数f(x,y)(x_0,y_0)偏导数的定义

\frac{\partial f(x,y)}{\partial 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}

\frac{\partial f(x,y)}{\partial 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}

在这个问题(x_0,y_0)=(0,0)
从而fx(0,0)的偏导数是

\lim_{\Delta x \rightarrow 0} \frac{f(0 + \Delta x, 0) – f(0,0)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{0 – 0}{\Delta x} = 0

从而fy(0,0)的偏导数是

\lim_{\Delta y \rightarrow 0} \frac{f(0, 0+\Delta y) – f(0,0)}{\Delta y} = \lim_{\Delta y \rightarrow 0}\sin\frac{1}{\Delta^2 y}

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