Use part I of the Fundamental Theorem of Calculus to find the derivative of:

F(x) = ( b = 4 , a = x) sin(t^3)dt

Here is part I of the fundamental Theorem of Calculus:

g(x) = (b = x, a = a) f(t)dt a<- x <- b is continuous on a, b g'(x) = f(x)

chain Rule

Part I of the Fundamental Theorem of Calculus says that if F(x) = (b , a = x) f(t)dt, then F'(x)= f(x). This just means replace t with x and you are done!!!

So for F(x) = ( b = 4 , a = x) sin(t^3)dt, then F'(x)= sin(x^3). It is that easy

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