Can you please check my work
Can you please tell me where I went wrong becasue my teacher said that answer is 13720
$\displaystyle f(x) = 400 e^{kx}$ where $\displaystyle k = \frac{1}{2} \ln (2.375)$.
$\displaystyle f'(x) = 400 k e^{kx}$.
$\displaystyle f'(10) = 400 k e^{10k}$.
$\displaystyle 10 k = 5 \ln (2.375) = \ln (2.375)^5 \Rightarrow e^{10k} = 2.375^5$.
So $\displaystyle f'(10) = 200 [\ln (2.375)] \, 2.375^5$.
Now use a calculator if a decimal approximation is required.