Question
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What is the derivative of $f(x) = x^{7} e^{0.2 x}$, evaluated at $x = 0.9$?

Solution
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Using the product rule we obtain
$$ f'(x) = e^{0.2 x} \cdot (7 \cdot x^6 + 0.2 x^7). $$
Evaluated at $x = 0.9$ and rounded to two digits the answer is
$f'(0.9) = 4.568278 = 4.57$.

Meta-information
================
extype: num
exsolution: 4.57
exname: exp derivative
extol: 0.01
