   \textbf{2.} Spo�t�te parci�ln� derivace 1. a 2. ��du pro funkci $z=z(x,y)$, je-li:
      $$
            x^2 + y^2 + z^2 = a^2.
      $$

      \textbf{�e�en�} Definujeme funkci $F(x,y,z) \equiv x^2+y^2+z^2-a^2.$ Tato funkce je
      t��dy~$\cal C^{\infty}.$ Parci�ln� derivace $\frac{\partial F}{\partial z} = 2z \neq 0,$ pro
      $z\neq0.$ V bodech, kde $z\neq0$ m��eme pou�it v�tu o implicitn� funkci:
      \begin{align*}
            x^2 + y^2 + z^2 - a^2&= 0\\
      \intertext{Derivujeme podle $x$:}
            2x + 2z\cdot\frac{\partial z}{\partial x}& = 0\\
            \frac{\partial z}{\partial x} &= -\frac xz\\
      \intertext{Ze symetrie $x$ a $y$ dost�v�me}
            \frac{\partial z}{\partial y} &= -\frac yz\\
      \intertext{Derivace druh�ho ��du:}
            \frac{\partial^2 z}{\partial y\partial x} & =
            \frac{\partial}{\partial y}\cdot\left(-\frac xz\right) =
            -\left(\frac{0-\frac{\partial z}{\partial y}\cdot x}{z^2}\right) =
            \frac{\partial z}{\partial y}\cdot\frac x{z^2} = -\frac{xy}{z^3}\\
            \frac{\partial^2 z}{\partial x^2} &=
            \frac{\partial}{\partial x}\left(-\frac xz\right) =
            -\frac{1\cdot z - \frac{\partial z}{\partial x}\cdot x}{z^2} =
            -\frac{z + x^2/z}{z^2} = -\frac{x^2+z^2}{z^3}
      \intertext{Ze symetrie $x$ a $y$}
            \frac{\partial^2 z}{\partial x^2} = -\frac{y^2+z^2}{z^3}.
      \end{align*}
