Question

derivative of ∫(-2,sin(x)) (cos(t^5)+t) dt

Answer 1

\[\int_{a}^{b}f(t)~dt=F(b)-F(a)\]
\[\frac{d}{dx}\left[\int_{a}^{b}f(t)~dt\right]=\frac{d}{dx}\left[F(b)-F(a)\right]\]
\[\frac{d}{dx}\left[\int_{a}^{b}f(t)~dt\right]=\frac{d}{dx}[F(b)]-\frac d{dx}F(a)]\]
\[\frac{d}{dx}\left[\int_{a}^{b}f(t)~dt\right]=f(b)b'-f(a)a'\]

Answer 2

since this is a general setup, it proves that we can go from the top to the bottom; so we really dont have to worry about the middle stuff ....

Answer 3

since the derivative of -2 is 0 ... all it amounts to really is:\[f(b)b'\]

Answer 4

That makes sense, thanks!