   \textbf{5.} Spo�t�te tot�ln� diferenci�l slo�en� funkce $u = f(x+y,z).$

      \textbf{�e�en�} Nech� $u = f(a, b)$, $a = a(x,y,z) = x+y$, $b = b(x,y,z) = z.$
      Tot�ln� diferenci�l je line�rn� zobrazen�, jeho� matice je tvo�ena parci�ln�mi
      derivacemi funkce $f.$ Spo�teme je pomoc� �et�zov�ho pravidla:
      \begin{align*}
            \frac{\partial u}{\partial x} &=
            \frac{\partial u}{\partial a} \cdot \frac{\partial a}{\partial x} +
            \frac{\partial u}{\partial b} \cdot \frac{\partial b}{\partial x} =
            \partial_1u\\
            \frac{\partial u}{\partial y} &=
            \frac{\partial u}{\partial a} \cdot \frac{\partial a}{\partial y} +
            \frac{\partial u}{\partial b} \cdot \frac{\partial b}{\partial y} =
            \partial_1u\\
            \frac{\partial u}{\partial z} &=
            \frac{\partial u}{\partial a} \cdot \frac{\partial a}{\partial z} +
            \frac{\partial u}{\partial b} \cdot \frac{\partial b}{\partial z} =
            \partial_2u
      \end{align*}
      Matice diferenci�lu m� tedy tvar
      $$
            u' = \begin{pmatrix} \partial_1u & \partial_1u & \partial_2u \end{pmatrix}
      $$ a diferenci�l je $\textrm{D}u = x\cdot\partial_1u + y\cdot\partial_1u +
      z\cdot\partial_2u.$
