 Zderivujte:
   $$
      y=\frac{{(1-x)}^p}{{(1+x)}^q}, \quad x\neq-1.
   $$
   \reseni
   Pou�ijeme k derivov�n� v�tu o derivaci pod�lu:
   \begin{equation*}\begin{split}
      y'&={\left(  \frac{{(1-x)}^p}{{(1+x)}^q}  \right)}'
         =\frac{{\bigl({(1-x)}^p\bigr)}'{(1+x)}^q -
                {\bigl({(1+x)}^q\bigr)}'{(1-x)}^p}
               {{(1+x)}^{2q}}=\\
        &=\frac{-p{(1-x)}^{p-1}{(1+x)}^q-q{(1+x)}^{q-1}{(1-x)}^p}
               {{(1+x)}^{2q}}=\\
        &=-\frac{{(1-x)}^{p-1}\cdot{(1+x)}^{q-1}\bigl(p(1+x) +
                q(1-x)\bigr)}{{(1+x)}^{2q}}=\\
        &=-\frac{{(1-x)}^{p-1}\cdot(p+px+q-qx)}
                {{(1+x)}^{q+1}}=
          -\frac{{(1-x)}^{p-1}}{{(1+x)}^{q+1}}\cdot\bigl((p-q)x+(p+q)\bigr).
   \end{split}\end{equation*}
