Answer: -1.5e^(-3/28), or -1.3476.
Explanation:
First of all, we need to determine in what term the derivative was taken so that we know what value should be substituted with 3 in the given derivative equation.
And if the derivative of 57 + 42e^(-t/28) is done in terms of t, using chain rule:
let,




Then, f(g(x)) = 42e^(-t/28), our initial expression.
According to chain rule, because
d/dx f(g(x)) = f'(g(x))*g'(x),
when substituting,
we get 42e^(-t/28)*(-1/28)
= -1.5e^(-t/28).
which is the first derivative of 57+42e^(-t/28) given in the question.
Therefore, we can conclude that the derivative of 57 + 42e^(-t/28) in terms of t is equal to -1.5e^(-t/28), and in the given equation
f’(3)= -1.5e^(-t/28).
t = 3.
Simple Substituting, we get
-1.5e^(-3/28), or -1.3476.
Feel free to leave a comment if you have any question!