Q: let f(x)=integral of e^(-t^2) limit from sin(x) to cos(x).Then find f'(pi/4)=
$\displaystyle f(x) = \int_{\sin x}^{\cos x} e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt + \int_{\sin x}^a e^{-t^2}\,dt = \int_a^{\cos x} e^{-t^2}\,dt - \int_a^{\sin x} e^{-t^2}\,dt$ so
$\displaystyle f'(x) = - e^{-\cos^2 x}\sin x - e^{-\sin^2 x}\cos x$
Then sub. in your value.
you don't need to solve the integral ... use the 2nd FTC.
$\displaystyle f(x) = \int_{\sin{x}}^{\cos{x}} e^{-t^2} \, dt$
$\displaystyle f'(x) = -e^{-\cos^2{x}} \sin{x} - e^{-\sin^2{x}} \cos{x}$
$\displaystyle f'\left(\frac{\pi}{4}\right) = -e^{-\frac{1}{2}} \cdot \frac{\sqrt{2}}{2} - e^{-\frac{1}{2}} \cdot \frac{\sqrt{2}}{2}
$
$\displaystyle f'\left(\frac{\pi}{4}\right) = -\sqrt{\frac{2}{e}}
$
let $\displaystyle \int_a^x f(t) \ dt = g(x),$ where $\displaystyle a$ is a constant. by the fundamental theorem of calculus we have $\displaystyle g'(x) = f(x).$ now if $\displaystyle u(x),v(x)$ are differentiable functions, then:
$\displaystyle h(x)=\int_{u(x)}^{v(x)} f(t) \ dt = g(v(x)) - g(u(x)).$ so by the chain rule: $\displaystyle h'(x)=v'(x)g'(v(x)) - u'(x)g'(u(x))=v'(x)f(v(x))-u'(x)f(u(x)).$
By making sure you integrate in an order that makes the problem tractable:
$\displaystyle \int_{y = 0}^1 \int_{x = 0}^{x = + \sqrt{y}} x e^{y^2} \, dx \, dy$.
There is a world of difference between double integrals and single integrals.
(I can't resist adding that you should have asked danny "How on erf do you do it?" .... Memo to baldeagle: I am the funniest )