$
\eqalign{
& f(x) = \sqrt x \cr
& f(x) = x^{\frac{1}
{2}} \cr
& f'(x) = \frac{1}
{2}x^{ - \frac{1}
{2}} \cr
& f'(x) = \frac{1}
{{2\sqrt x }} \cr}
$
$
\eqalign{
& f(x) = \sqrt {3x - 4} \cr
& f(x) = \left( {3x - 4} \right)^{\frac{1}
{2}} \cr
& f'(x) = \frac{1}
{2}\left( {3x - 4} \right)^{ - \frac{1}
{2}} \cdot 3 \cr
& f'(x) = \frac{3}
{{2\sqrt {3x - 4} }} \cr}
$
Hoe bereken je afgeleide van $f(x)=x\sqrt{8-x^2}$?
$
\eqalign{
& f(x) = x\sqrt {8 - x^2 } \cr
& f'(x) = \sqrt {8 - x^2 } + x \cdot \frac{1}
{{2\sqrt {8 - x^2 } }} \cdot - 2x \cr
& f'(x) = \sqrt {8 - x^2 } - \frac{{x^2 }}
{{\sqrt {8 - x^2 } }} \cr
& f'(x) = \frac{{8 - x^2 }}
{{\sqrt {8 - x^2 } }} - \frac{{x^2 }}
{{\sqrt {8 - x^2 } }} \cr
& f'(x) = \frac{{8 - 2x^2 }}
{{\sqrt {8 - x^2 } }} \cr}
$