1.
\eqalign{ & f(x) = \sqrt {\frac{x} {{x - 1}}} \cr & f'(x) = \frac{1} {{2 \cdot \sqrt {\frac{x} {{x - 1}}} }} \cdot \frac{{ - 1}} {{\left( {x - 1} \right)^2 }} \cr & f'(x) = \frac{1} {{2 \cdot \sqrt {\frac{x} {{x - 1}}} }} \cdot \frac{{ - 1}} {{\left( {x - 1} \right)^2 }} \cdot \frac{{\sqrt {\frac{x} {{x - 1}}} }} {{\sqrt {\frac{x} {{x - 1}}} }} \cr & f'(x) = \frac{1} {{2 \cdot \frac{x} {{x - 1}}}} \cdot \frac{{ - 1\sqrt {\frac{x} {{x - 1}}} }} {{\left( {x - 1} \right)^2 }} \cr & f'(x) = \frac{1} {{2x}} \cdot \frac{{ - 1\sqrt {\frac{x} {{x - 1}}} }} {{x - 1}} \cr & f'(x) = - \frac{{\sqrt {\frac{x} {{x - 1}}} }} {{2x\left( {x - 1} \right)}} \cr}
2.
\eqalign{ & f(x) = 1 + \sqrt {x^2 - x} \cr & f'(x) = \frac{1} {{2\sqrt {x^2 - x} }} \cdot \left( {2x - 1} \right) = \frac{{2x - 1}} {{2\sqrt {x^2 - x} }} \cr & f''(x) = \frac{{2 \cdot 2\sqrt {x^2 - x} - \left( {2x - 1} \right) \cdot \frac{{2x - 1}} {{\sqrt {x^2 - x} }}}} {{\left( {2\sqrt {x^2 - x} } \right)^2 }} \cr & f''(x) = \frac{{4\sqrt {x^2 - x} - \frac{{\left( {2x - 1} \right)^2 }} {{\sqrt {x^2 - x} }}}} {{\left( {2\sqrt {x^2 - x} } \right)^2 }} \cr & f''(x) = \frac{{\frac{{4\sqrt {x^2 - x} \cdot \sqrt {x^2 - x} }} {{\sqrt {x^2 - x} }} - \frac{{\left( {2x - 1} \right)^2 }} {{\sqrt {x^2 - x} }}}} {{4\left( {\sqrt {x^2 - x} } \right)^2 }} \cr & f''(x) = \frac{{\frac{{4\left( {x^2 - x} \right) - \left( {2x - 1} \right)^2 }} {{\sqrt {x^2 - x} }}}} {{4\left( {\sqrt {x^2 - x} } \right)^2 }} \cr & f''(x) = \frac{{4\left( {x^2 - x} \right) - \left( {2x - 1} \right)^2 }} {{4\left( {\sqrt {x^2 - x} } \right)^3 }} \cr & f''(x) = \frac{{ - 1}} {{4\left( {\sqrt {x^2 - x} } \right)^3 }} \cr & f''(x) = - \frac{1} {{4\left( {\sqrt {x^2 - x} } \right)^3 }} \cr}
3.
f(x) = \large\frac{4}{{\sqrt {2x^3 + 5x^2 + 4} }}
f(x) = 4\left( {2x^3 + 5x^2 + 4} \right)^{ - \frac{1}{2}}
Dan de afgeleide:
f'(x) = 4 \cdot - \frac{1}{2}\left( {2x^3 + 5x^2 + 4} \right)^{ - 1\frac{1}{2}} \cdot \left( {6x^2 + 10x} \right)
f'(x) = - \large\frac{{2\left( {6x^2 + 10x} \right)}}{{\sqrt {\left( {2x^3 + 5x^2 + 4} \right)^3 } }}
f'(x) = - \large\frac{{12x^2 + 20x}}{{\sqrt {\left( {2x^3 + 5x^2 + 4} \right)^3 } }}
