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I would like some guidance on how to solve these type of problems.

Find $h'(x)$ if $$h(x) = \int\limits_{\cos(x)}^x \mathrm{e}^{t^2} \, dt$$

Mathematica says $h'(x) = e^{x^2} - e^{\cos^2(x)} ( - \sin x )$

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2 Answers 2

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This is general formula for differentiating a integral wrt. x

$$(d/dx)\int\limits_{h \left(x\right)}^{g\left(x\right)} F\left(t,x\right) . dt =g'(x).F\left(g(x),x\right) - h'(x).F(h(x),x) + \int\limits_{h(x)}^{g(x)} F'(t,x).dt $$

$$g'(x) , h'(x) , F'(t,x) $$ are the partial differentials wrt. x[keeping t a constant].

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@shuxiaoli your method is not general , ie if $F()$ is a function of x and t both then we have to use the generalised form. –  Mr.ØØ7 Mar 20 '13 at 2:25

Let $F'(t)=e^{t^2}$. Then, $h(x)=F(x)-F(\cos x)$.

Therefore, $h'(x)=F'(x)-F'(\cos x)=e^{x^2}+e^{\cos^2x}\sin x.$

Don't forget the Chain Rule.

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Yeah, I was going to tell you to not forget the Chain Rule :-) I see the edit now. –  Patrick Mar 20 '13 at 0:17

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