deriv: Product Rule for Derivatives

Using the product rule for \(f(x) = g(x) \cdot h(x)\), where \(g(x) := x^{7}\) and \(h(x) := e^{3.9 x}\), we obtain \[ \begin{aligned} f'(x) &= [g(x) \cdot h(x)]' = g'(x) \cdot h(x) + g(x) \cdot h'(x) \\ &= 7 x^{7 - 1} \cdot e^{3.9 x} + x^{7} \cdot e^{3.9 x} \cdot 3.9 \\ &= e^{3.9 x} \cdot(7 x^6 + 3.9 x^{7}) \\ &= e^{3.9 x} \cdot x^6 \cdot (7 + 3.9 x). \end{aligned} \] Evaluated at \(x = 0.51\), the answer is \[ e^{3.9 \cdot 0.51} \cdot 0.51^6 \cdot (7 + 3.9 \cdot 0.51) = 1.155964. \] Thus, rounded to two digits we have \(f'(0.51) = 1.16\).

Using the product rule for \(f(x) = g(x) \cdot h(x)\), where \(g(x) := x^{5}\) and \(h(x) := e^{3.1 x}\), we obtain \[ \begin{aligned} f'(x) &= [g(x) \cdot h(x)]' = g'(x) \cdot h(x) + g(x) \cdot h'(x) \\ &= 5 x^{5 - 1} \cdot e^{3.1 x} + x^{5} \cdot e^{3.1 x} \cdot 3.1 \\ &= e^{3.1 x} \cdot(5 x^4 + 3.1 x^{5}) \\ &= e^{3.1 x} \cdot x^4 \cdot (5 + 3.1 x). \end{aligned} \] Evaluated at \(x = 0.72\), the answer is \[ e^{3.1 \cdot 0.72} \cdot 0.72^4 \cdot (5 + 3.1 \cdot 0.72) = 18.110635. \] Thus, rounded to two digits we have \(f'(0.72) = 18.11\).

Using the product rule for \(f(x) = g(x) \cdot h(x)\), where \(g(x) := x^{3}\) and \(h(x) := e^{4 x}\), we obtain \[ \begin{aligned} f'(x) &= [g(x) \cdot h(x)]' = g'(x) \cdot h(x) + g(x) \cdot h'(x) \\ &= 3 x^{3 - 1} \cdot e^{4 x} + x^{3} \cdot e^{4 x} \cdot 4 \\ &= e^{4 x} \cdot(3 x^2 + 4 x^{3}) \\ &= e^{4 x} \cdot x^2 \cdot (3 + 4 x). \end{aligned} \] Evaluated at \(x = 0.56\), the answer is \[ e^{4 \cdot 0.56} \cdot 0.56^2 \cdot (3 + 4 \cdot 0.56) = 15.435723. \] Thus, rounded to two digits we have \(f'(0.56) = 15.44\).