Final answer:
To find f '(0) for the function f(x) = ln(x + 4 + e^(-3x)), we take the derivative of the function with respect to x and evaluate it at x = 0.
Step-by-step explanation:
To find f '(0) for the function f(x) = ln(x + 4 + e^(-3x)), we need to take the derivative of the function with respect to x and then evaluate it at x = 0.
To find the derivative, we use the chain rule. The derivative of ln(u) = (1/u) * du/dx. Applying this rule to our function, we have:
f '(x) = (1/(x + 4 + e^(-3x))) * ((1/(x + 4 + e^(-3x))) * (1 + e^(-3x)) + 3e^(-3x))
Evaluating f '(x) at x = 0 gives:
f '(0) = (1/(0 + 4 + e^0)) * ((1/(0 + 4 + e^0)) * (1 + e^0) + 3e^0)