// Numbas version: finer_feedback_settings
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a) The derivative of \\[ f(x) = \\sqrt[\\var{n}]{x} = x^{\\frac{1} {\\var{n}}} \\] is given by
\\[ f'(x) = \\frac{\\mathrm{d}}{\\mathrm{d}x}(x^{\\frac{1} {\\var{n}}}) = {\\frac{1} {\\var{n}}} \\cdot x^{\\frac{\\var{1-n}} {\\var{n}}} .\\]
The equation of the tangent to the graph of \\( f \\) at \\( ( a, f(a)) \\) is obtained by
\\[ y - f(a) = f'(a) \\cdot (x-a) \\]
For \\( a = \\var{a} \\) , this means
\\[ y - \\var{m} = \\frac{1} {\\var{ricoN}} \\cdot \\left( x-\\var{a} \\right) \\]
or equivalently
\\[ y = \\var{m} + \\frac{1} {\\var{ricoN}} \\cdot \\left( x-\\var{a} \\right) = \\frac{1} {\\var{ricoN}} \\cdot x + \\frac {\\var{m*(n-1)}} {\\var{n}}\\]
b) The linear approximation of \\( f(a+\\var{eps}) \\) is given by
\\[ f(a) + f'(a) \\cdot \\var{eps} \\]
When \\( a = \\var{a} \\) this leads to
\\[ \\var{rklben} \\]
which needs to be rounded up to \\( \\var{p+3} \\) decimals.
cst rklvgl noemer
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\nDo this in the form \\( y = number1 \\cdot x+number2 \\). Use fractions, but no decimals.
\n\\( y = \\) [[0]] \\( \\cdot x+ \\) [[1]]
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\n\\( f(\\var{a+eps}) \\) approximately equals [[0]]
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