Текст задатака објашњених у видео лекцији:

Пр.3)   Одредити други извод функције:

            $y = \frac{{2{x^2} - 2x + 1}}{{x - 1}}$

Пр.4)   $y = \ln \sqrt {1 - {x^2}} $

Пр.5)   $y = arctg\frac{x}{{\sqrt {1 - {x^2}} }}$

Пр.3)   $y = \frac{{2{x^2} - 2x + 1}}{{x - 1}}$

\[\begin{gathered}
y' = \frac{{{{\left( {2{x^2} - 2x + 1} \right)}^\prime }\left( {x - 1} \right) - {{\left( {x - 1} \right)}^\prime }\left( {2{x^2} - 2x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \hfill \\
= \frac{{\left( {4x - 2} \right)\left( {x - 1} \right) - \left( {2{x^2} - 2x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \hfill \\
= \frac{{4{x^2} - 4x - 2x + 1 - 2{x^2} + 2x - 1}}{{{{\left( {x - 1} \right)}^2}}} = \frac{{2{x^2} - 4x + 1}}{{{{\left( {x - 1} \right)}^2}}} \hfill \\
\end{gathered} \]

\[\begin{gathered}
y'' = \frac{{{{\left( {2{x^2} - 4x + 1} \right)}^\prime }{{\left( {x - 1} \right)}^2} - \left( {2{x^2} - 4x + 1} \right){{\left( {{{\left( {x - 1} \right)}^2}} \right)}^\prime }}}{{{{\left( {x - 1} \right)}^4}}} = \hfill \\
= \frac{{\left( {4x - 4} \right){{\left( {x - 1} \right)}^2} - \left( {2{x^2} - 4x + 1} \right)2\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^4}}} = \hfill \\
= \frac{{\left( {x - 1} \right)\left( {\left( {4x - 4} \right)\left( {x - 1} \right) - \left( {2{x^2} - 4x + 1} \right)2} \right)}}{{{{\left( {x - 1} \right)}^4}}} = \hfill \\
= \frac{{\left( {4{x^2} - 4x - 4x + 4 - 4{x^2} + 8x - 2} \right)}}{{{{\left( {x - 1} \right)}^3}}} = \frac{2}{{{{\left( {x - 1} \right)}^3}}} \hfill \\
\end{gathered} \]

Пр.4)   $y = \ln \sqrt {1 - {x^2}} $

\[\begin{gathered}
y' = \frac{1}{{\sqrt {1 + {x^2}} }}{\left( {\sqrt {1 + {x^2}} } \right)^\prime } = \frac{1}{{\sqrt {1 + {x^2}} }}{\left( {{{\left( {1 + {x^2}} \right)}^{\frac{1}{2}}}} \right)^\prime } = \hfill \\
= \frac{1}{{\sqrt {1 + {x^2}} }}\frac{1}{2}{\left( {1 + {x^2}} \right)^{\frac{1}{2} - 1}}{\left( {1 + {x^2}} \right)^\prime } = \frac{{2x}}{{2\sqrt {1 + {x^2}} \sqrt {1 + {x^2}} }} = \frac{x}{{1 + {x^2}}} \hfill \\
y'' = \frac{{x'\left( {1 + {x^2}} \right) - {{\left( {1 + {x^2}} \right)}^\prime }x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{1 + {x^2} - 2x \cdot x}}{{{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{1 - {x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \hfill \\
\end{gathered} \]

Пр.5)   $y = arctg\frac{x}{{\sqrt {1 - {x^2}} }}$

\[\begin{gathered}
y' = \frac{1}{{1 + {{\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)}^2}}}{\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)^\prime } = \frac{1}{{1 + \frac{{{x^2}}}{{1 - {x^2}}}}}\left( {\frac{{x'\sqrt {1 - {x^2}} - {{\left( {\sqrt {1 - {x^2}} } \right)}^\prime }x}}{{{{\left( {\sqrt {1 - {x^2}} } \right)}^2}}}} \right) = \hfill \\
= \frac{1}{{1 + \frac{{{x^2}}}{{1 - {x^2}}}}}\left( {\frac{{x'\sqrt {1 - {x^2}} - {{\left( {\sqrt {1 - {x^2}} } \right)}^\prime }x}}{{{{\left( {\sqrt {1 - {x^2}} } \right)}^2}}}} \right) = \hfill \\
= \frac{1}{{\frac{{1 - {x^2} + {x^2}}}{{1 - {x^2}}}}} \cdot \frac{{\sqrt {1 - {x^2}} - x\frac{1}{2}{{\left( {1 - {x^2}} \right)}^{ - \frac{1}{2}}}{{\left( {1 - {x^2}} \right)}^\prime }}}{{1 - {x^2}}} = \hfill \\
= \frac{{1 - {x^2}}}{1} \cdot \frac{{\sqrt {1 - {x^2}} - x\frac{1}{2}{{\left( {1 - {x^2}} \right)}^{ - \frac{1}{2}}}\left( { - 2x} \right)}}{{1 - {x^2}}} = \hfill \\
= \sqrt {1 - {x^2}} + \frac{{{x^2}}}{{\sqrt {1 - {x^2}} }} = \frac{{1 - {x^2} + {x^2}}}{{\sqrt {1 - {x^2}} }} = \frac{1}{{\sqrt {1 - {x^2}} }} \hfill \\
\end{gathered} \]

\[y'' =  - \frac{1}{2}{\left( {1 - {x^2}} \right)^{ - \frac{3}{2}}}{\left( {1 - {x^2}} \right)^\prime } =  - \frac{{ - 2x}}{{2{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}}} = \frac{x}{{{{\sqrt {\left( {1 - {x^2}} \right)} }^3}}}\]