What is the derivative of \(f(x) = x^{8} e^{3.4x}\), evaluated at \(x = 0.7\)?
Using the product rule for \(f(x) = g(x) \cdot h(x)\), where \(g(x) := x^{8}\) and \(h(x) := e^{3.4x}\), we obtain \[\begin{aligned} f'(x) & = & [g(x) \cdot h(x)]' = g'(x) \cdot h(x) + g(x) \cdot h'(x) \\ & = & 8 x^{8 - 1} \cdot e^{3.4x} + x^{8} \cdot e^{3.4x} \cdot 3.4 \\ & = & e^{3.4x} \cdot(8 x^7 + 3.4 x^{8}) \\ & = & e^{3.4x} \cdot x^7 \cdot (8 + 3.4x).\end{aligned}\] Evaluated at \(x = 0.7\), the answer is \[e^{3.4\cdot 0.7} \cdot 0.7^7 \cdot (8 + 3.4\cdot 0.7) = 9.236438.\] Thus, rounded to two digits we have \(f'(0.7) = 9.24\).