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$$\large f\left(x\right) = \int\limits_{\cos x}^{\sin x} e^{t^2+xt}dt.$$Compute $f'\left(0\right)$.

I can't get it right -sigh- :/

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If $x=0$ then the integral is..... –  N. S. Jun 20 '12 at 19:11
    
Sorry @N.S. it is supposed to be first derivative of the function $f(x)$. I have edited the question now. –  MSC Jun 20 '12 at 19:14
    
Use leibniz integral rule . –  Theorem Jun 20 '12 at 19:16
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up vote 5 down vote accepted

If we have $$f(x) = \int_{a(x)}^{b(x)} g(t,x) dt,$$ then for "nice enough" $g(t,x)$ $$f'(x) = \int_{a(x)}^{b(x)} \dfrac{\partial g(t,x)}{\partial x} dt + g(b(x),x) \dfrac{db(x)}{dx} - g(a(x),x) \dfrac{da(x)}{dx}$$ In your case, $g(t,x) = \exp(t^2 + xt)$, $a(x) = \cos(x)$ and $b(x) = \sin(x)$.

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$$\dfrac{\partial g(t,x)}{\partial x} = t\exp(t^2+xt), \dfrac{db(x)}{dx} = \cos(x), \dfrac{da(x)}{dx} = -\sin(x)$$ Hence, $$f'(0) = \int_{a(0)}^{b(0)} t\exp(t^2) dt + g(b(0),0) \cos(0) - g(a(0),0) (-\sin(0))$$ $$ = \int_{1}^{0} t\exp(t^2) dt + g(0,0) \times 1 + g(1,0) \times 0$$ $$ = \left. \dfrac{\exp(t^2)}{2} \right \vert_1^0 + 1 = 1 + \left(\dfrac{1 - \exp(1)}2 \right) = \left(\dfrac{3 - \exp(1)}2 \right)$$

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You can probably handle it yourself, after a brief look at this article about differentiating under the integral sign.

For $f'(x)$, you will get some easy terms, and the unpleasant-looking $$\int_{\cos x}^{\sin x} te^{t^2+xt}\,dt.$$ However, for $x=0$, this integral is very tame.

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