Antwoorden
Oefening 1
\eqalign{ & f(x) = \frac{{x^2 - 5}} {{2x - 4}} \cr & f'(x) = \frac{{2x \cdot \left( {2x - 4} \right) - \left( {x^2 - 5} \right) \cdot 2}} {{\left( {2x - 4} \right)^2 }} = \frac{{4x^2 - 8x - 2x^2 + 10}} {{\left( {2x - 4} \right)^2 }} = \cr & = \frac{{2x^2 - 8x + 10}} {{\left( {2x - 4} \right)^2 }} = \frac{{2\left( {x^2 - 4x + 5} \right)}} {{4\left( {x - 2} \right)^2 }} = \frac{{x^2 - 4x + 5}} {{2\left( {x - 2} \right)^2 }} \cr}
Zie ook De afgeleide van f(x)=(x2-5)/(2x-4)
Oefening 2
\eqalign{ & f(x) = \frac{{x^3 \ln x}} {{x + 2}} \cr & f'(x) = \frac{{\left( {3x^2 \ln x + x^3 \cdot \frac{1} {x}} \right) \cdot \left( {x + 2} \right) - x^3 \ln x \cdot 1}} {{\left( {x + 2} \right)^2 }} \cr & f'(x) = \frac{{3x^3 \ln x + x^3 + 6x^2 \ln x + 2x^2 - x^3 \ln x}} {{\left( {x + 2} \right)^2 }} \cr & f'(x) = \frac{{2x^3 \ln x + x^3 + 6x^2 \ln x + 2x^2 }} {{\left( {x + 2} \right)^2 }} \cr & f'(x) = \frac{{x^2 \left( {2x\ln x + 6\ln x + x + 2} \right)}} {{\left( {x + 2} \right)^2 }} \cr & f'(x) = \frac{{x^2 \left( {\left( {2x + 6} \right)\ln x + x + 2} \right)}} {{\left( {x + 2} \right)^2 }} \cr}
Oefening 3
\eqalign{ & h(t) = \left( {\frac{{\sqrt t }} {{4 + t^5 }}} \right)^{10} = \frac{{t^5 }} {{\left( {4 + t^5 } \right)^{10} }} \cr & h'(t) = \frac{{5t^4 \cdot \left( {4 + t^5 } \right)^{10} - t^5 \cdot 10\left( {4 + t^5 } \right)^9 \cdot 5t^4 }} {{\left( {4 + t^5 } \right)^{20} }} \cr & h'(t) = \frac{{5t^4 \cdot \left( {4 + t^5 } \right) - t^5 \cdot 10 \cdot 5t^4 }} {{\left( {4 + t^5 } \right)^{11} }} \cr & h'(t) = \frac{{20t^4 + 5t^9 - 50t^9 }} {{\left( {4 + t^5 } \right)^{11} }} \cr & h'(t) = \frac{{20t^4 - 45t^9 }} {{\left( {4 + t^5 } \right)^{11} }} \cr & En\,\,dan\,\,eventueel: \cr & h'(t) = \frac{{5t^4 \left( {4 - 9t^5 } \right)}} {{\left( {4 + t^5 } \right)^{11} }} \cr}
Oefening 4
\eqalign{ & k(x) = \frac{{4x - 7}} {{3 - x^2 }} \cr & k'(x) = \frac{{4 \cdot \left( {3 - x^2 } \right) - \left( {4x - 7} \right) \cdot - 2x}} {{\left( {3 - x^2 } \right)^2 }} \cr & k'(x) = \frac{{12 - 4x^2 + 8x^2 - 14x}} {{\left( {3 - x^2 } \right)^2 }} \cr & k'(x) = \frac{{4x^2 - 14x + 12}} {{\left( {3 - x^2 } \right)^2 }} \cr}
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