Exam 1

  1. Question

    What is the derivative of f(x)=x8e3.4xf(x) = x^{8} e^{3.4 x}, evaluated at x=0.7x = 0.7?


    Solution

    Using the product rule for f(x)=g(x)h(x)f(x) = g(x) \cdot h(x), where g(x):=x8g(x) := x^{8} and h(x):=e3.4xh(x) := e^{3.4 x}, we obtain f(x)=[g(x)h(x)]=g(x)h(x)+g(x)h(x)=8x81e3.4x+x8e3.4x3.4=e3.4x(8x7+3.4x8)=e3.4xx7(8+3.4x). \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^{3.4 x} + x^{8} \cdot e^{3.4 x} \cdot 3.4 \\ &= e^{3.4 x} \cdot(8 x^7 + 3.4 x^{8}) \\ &= e^{3.4 x} \cdot x^7 \cdot (8 + 3.4 x). \end{aligned} Evaluated at x=0.7x = 0.7, the answer is e3.40.70.77(8+3.40.7)=9.236438. 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)=9.24f'(0.7) = 9.24.