Using the Fundamental theorem of Calculus find the derivative of

F(x) = sec(3x) + ∫ [e^-(t^(2))]
Integral from ( 0 to tan(3x) )

f(x) = sec(x)
f'(x)= sec(x)tan(x)
and according to the fundamental ToC, u plug the x value of the integral into the the t value of the function.
Would the answer be
f'(x) = 3sec(3x)tan(3x) + e^(-tan(3x)^(2)

almost. It should be

f'(x) = 3sec(3x)tan(3x) + e^(-tan^2(3x)) * 3sec^2(3x)
it's just the chain rule

To use the Fundamental Theorem of Calculus to find the derivative of F(x), you need to evaluate the derivative of the integrand and then substitute the upper limit of the integral into the result.

Let's break down the problem step by step:

1. Given function: F(x) = sec(3x) + ∫ [e^-(t^2)] from 0 to tan(3x)
2. First, find the derivative of the integral:
∫ [e^-(t^2)] = F(t) (we'll use F(t) to differentiate the integral)
F'(t) = e^-(t^2)
3. Now, differentiate the first term of F(x): sec(3x)
f(x) = sec(x)
f'(x) = sec(x)tan(x) (use the derivative rule for sec(x))
4. Finally, substitute the upper limit, tan(3x), into the derivative of the integral:
f'(x) = 3sec(3x)tan(3x) + e^-(tan(3x)^2)

Therefore, the derivative of F(x) is:
F'(x) = 3sec(3x)tan(3x) + e^-(tan(3x)^2)