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