方向导数存在但是偏导数不存在的例子
\[\begin{gathered} z = \sqrt {{x^2} + {y^2}} \hfill \\ {\left. {\frac{{\partial z}}{{\partial l}}} \right|_{\left( {0,0} \right)}} = \hfill \\ \mathop {\lim }\limits_{t \to {0^ + }} \frac{{f\left( {{x_0} + t,{y_0}} \right) - f\left( {{x_0},{y_0}} \right)}}{t} \hfill \\ = \mathop {\lim }\limits_{t \to {0^ + }} \frac{{\sqrt {{{\left( {{x_0} + t} \right)}^2} + y_0^2} - \sqrt {{x_0}^2 + y_0^2} }}{t} \hfill \\ = \mathop {\lim }\limits_{t \to {0^ + }} \frac{{\sqrt {{t^2}} }}{t} \hfill \\ = 1 \hfill \\ {\left. {\frac{{\partial z}}{{\partial x}}} \right|_{\left( {0,0} \right)}} = \hfill \\ \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {{x_0} + \Delta x,{y_0}} \right) - f\left( {{x_0},{y_0}} \right)}}{{\Delta x}} \hfill \\ = \hfill \\ \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sqrt {{{\left( {{x_0} + \Delta x} \right)}^2} + y_0^2} - \sqrt {{x_0}^2 + y_0^2} }}{{\Delta x}} \hfill \\ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left| {\Delta x} \right|}}{{\Delta x}} = \left\{ \begin{gathered} \frac{{\Delta x}}{{\Delta x}} = 1,\Delta x > 0, \hfill \\ \frac{{ - \Delta x}}{{\Delta x}} = - 1,\Delta x < 0 \hfill \\ \end{gathered} \right. \hfill \\ \therefore {\left. {\frac{{\partial z}}{{\partial x}}} \right|_{\left( {0,0} \right)}}\begin{array}{*{20}{c}} {}&{not\begin{array}{*{20}{c}} {}&{exit} \end{array}} \end{array}. \hfill \\ \end{gathered} \]