$$F(x)=\int _{ 1 }^{ ln(x^{ 2 }+e) }{ sin(e^{ t }) } dt$$
Find F'(x).
Not sure how to do this, I understand from the Fundamental Theorem of Calculus that
$$F'(x)=\frac { d }{ dx } \int _{ a }^{ x }{ f(t) } dt=f(x)$$
But what happens to the limit of $ln(x^2+e)$?

Question

$$F(x)=\int _{ 1 }^{ ln(x^{ 2 }+e) }{ sin(e^{ t }) } dt$$
Find F'(x).
Not sure how to do this, I understand from the Fundamental Theorem of Calculus that 
$$F'(x)=\frac { d }{ dx } \int _{ a }^{ x }{ f(t) } dt=f(x)$$
But what happens to the limit of $ln(x^2+e)$?

bahrom7893 · Answer

Don't you just do this:
Sin(e^b)*(Sin(e^b)) ' where b is the upper limit.

bahrom7893 · Answer

Well, you'd then subtract Sin(e^1) * (Sin(e^1))', but since the derivative of a constant is 0, this term becomes 0, and you're left with just the first one.

bahrom7893 · Answer

Oh and by the way, e^(ln(x)) = x

phoenixfire · Answer

I don't understand why you're multiplying f(x) by f'(x)
Or do you do the integration, and then derive it?
$$\frac{d}{dx}(sin(e^{ln(x^2+e)})-sin(e^1))=\frac{d}{dx}sin(x^2+e)=2xcos(x^2+e)=F'(x)$$