Question

What is the instantaneous rate of change of `f(x)=e^(5x-7)` at `x=0`?

Answer 1

`5e^(-7)` = 0.00456, nearly.

Explanation:

`f' = 5 e^(3x-7)`, using `d/dx(e^u)=d/(du)(e^u)d/dx(u)`

At x = 0, f' = 5`e^(-7)`