What is the derivative of f(x)=x7e3.9xf(x) = x^{7} e^{3.9 x}, evaluated at x=0.51x = 0.51?
Using the product rule for f(x)=g(x)⋅h(x)f(x) = g(x) \cdot h(x), where g(x):=x7g(x) := x^{7} and h(x):=e3.9xh(x) := e^{3.9 x}, we obtain f′(x)=[g(x)⋅h(x)]′=g′(x)⋅h(x)+g(x)⋅h′(x)=7x7−1⋅e3.9x+x7⋅e3.9x⋅3.9=e3.9x⋅(7x6+3.9x7)=e3.9x⋅x6⋅(7+3.9x). \begin{aligned} f'(x) &= [g(x) \cdot h(x)]' = g'(x) \cdot h(x) + g(x) \cdot h'(x) \\ &= 7 x^{7 - 1} \cdot e^{3.9 x} + x^{7} \cdot e^{3.9 x} \cdot 3.9 \\ &= e^{3.9 x} \cdot(7 x^6 + 3.9 x^{7}) \\ &= e^{3.9 x} \cdot x^6 \cdot (7 + 3.9 x). \end{aligned} Evaluated at x=0.51x = 0.51, the answer is e3.9⋅0.51⋅0.516⋅(7+3.9⋅0.51)=1.155964. e^{3.9 \cdot 0.51} \cdot 0.51^6 \cdot (7 + 3.9 \cdot 0.51) = 1.155964. Thus, rounded to two digits we have f′(0.51)=1.16f'(0.51) = 1.16.