$
\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}
$
