Antwoorden

$
\eqalign{
  & f(x) = \sqrt x  = x^{\frac{1}
{2}}   \cr
  & f'(x) = \frac{1}
{2}x^{ - \frac{1}
{2}}  = \frac{1}
{2} \cdot \frac{1}
{{\sqrt x }} = \frac{1}
{{2\sqrt x }}  \cr
  &   \cr
  & g(x) = 2\root 3 \of x  = 2x^{\frac{1}
{3}}   \cr
  & g'(x) = 2 \cdot \frac{1}
{3} \cdot x^{ - \frac{2}
{3}}  = \frac{2}
{3} \cdot \frac{1}
{{\root 3 \of {x^2 } }} = \frac{2}
{{3\root 3 \of {x^2 } }}  \cr
  &   \cr
  & h(x) = x^3 \sqrt x  = x^{3\frac{1}
{2}}   \cr
  & h'(x) = 3\frac{1}
{2}x^{2\frac{1}
{2}}  = 3\frac{1}
{2}x^2  \cdot x^{\frac{1}
{2}}  = 3\frac{1}
{2}x^2 \sqrt x  \cr}
$ 

Nog meer oefeningen

$
\eqalign{
  & a.  \cr
  & f(x) = (x + \sqrt x )^2   \cr
  & f'(x) = 2 \cdot (x + \sqrt x ) \cdot \left( {1 + \frac{1}
{{2\sqrt x }}} \right)  \cr
  & b.  \cr
  & g(x) = \sqrt {(1 + x^2 )^3 }  = (1 + x^2 )^{1\frac{1}
{2}}   \cr
  & g'(x) = 1\frac{1}
{2}(1 + x^2 )^{\frac{1}
{2}}  \cdot 2x = 3x\sqrt {1 + x^2 }   \cr
  & c.  \cr
  & h(x) = \sqrt {(x + 2)^2  - (x - 2)^2 }  = \sqrt {x^2  + 4x + 4 - \left( {x^2  - 4x + 4} \right)}  = \sqrt {8x}   \cr
  & h'(x) = \frac{1}
{{2\sqrt {8x} }} \cdot 8 = \frac{4}
{{\sqrt {8x} }} = \frac{2}
{{\sqrt {2x} }} \cr}
$

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