 \textbf{1.} Spo�t�te parci�ln� derivaci podle prvn� prom�nn� v bod� $(0,1,1)$
      funkce $h: \mathbb R^3 \to \mathbb R,$ kter� vznikne slo�en�m funkce
      $g: \mathbb R^3 \to \mathbb R,$ jej� diferenci�l v bod� $(2,2,2)$ je
      reprezentov�n matic� $\begin{pmatrix}1&2&3\end{pmatrix}$
      a zobrazen� $f: \mathbb R^3 \to \mathbb R^3,$
      kter� je definov�no p�edpisem $$f(u,v,w) = (u+v+w,u^2+v^2+1, u^3+v^3+1).$$

      \textbf{�e�en�} Nech� $h = h(x, y, z)$. V�me, �e
      \begin{align*}
            x&= x(u,v,w) = u+v+w\\
            y&= y(u,v,w) = u^2+v^2+1\\
            z&= z(u,v,w) = u^3 + v^3 + 1
      \end{align*}
      a
      $$
            \frac{\partial h}{\partial x} = 1,\quad\frac{\partial h}{\partial y} = 2,
            \quad\frac{\partial h}{\partial z} = 3
      $$
      v bod� $(2,2,2)$
      Pot�ebujeme vypo��tat $\frac{\partial h}
      {\partial u} \bigl((0,1,1)\bigr).$
      K v�po�tu pou�ijeme �et�zov� pravidlo (parci�ln� derivace $x$, $y$ a $z$ jsou v
      bod� $(u,v,w) = (0,1,1)$ a parci�ln� derivace funkce $h$ je v bod� $(x,y,z) = (2,2,2)$):
      $$
            \frac{\partial h}{\partial u}\bigl((0,1,1)\bigr) =
            \frac{\partial h}{\partial x}\cdot\frac{\partial x}{\partial u} +
            \frac{\partial h}{\partial y}\cdot\frac{\partial y}{\partial u} +
            \frac{\partial h}{\partial z}\cdot\frac{\partial z}{\partial u} =
            1\cdot1 + 2\cdot0 + 3\cdot0 = 1.
      $$
