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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