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Suppose we have: $F(x)=\int_{a(x)}^{b(x)}e^{h(x,t)}dt$. Is it true that $F^{'}(x)=\int_{a(x)}^{b(x)}\frac{\partial h(x,t)}{\partial x}.e^{h(x,t)}dt$ ? Please tell me under what condition(s) am I allowed to use this?

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What you have is not quite right. I recommend you try using the Fundamental Theorem of Calculus. – Alex Becker Jan 26 '12 at 22:20
I think this might be what you are after. – Emilio Ferrucci Jan 26 '12 at 22:30
up vote 1 down vote accepted

No it's not true in general (it is true when $a(x)$ and $b(x)$ are constant). To see why let's come back to the definition and write down $$F(x+h)-F(x)=\int_{a(x+h)}^{b(x+h)}e^{h(x+h,t)}dt-\int_{a(x)}^{b(x)}e^{h(x,t)}dt$$ $$=\int_{a(x)}^{b(x)}e^{h(x+h,t)}dt-\int_{a(x)}^{b(x)}e^{h(x,t)}dt+\int_{b(x)}^{b(x+h)}e^{h(x+h,t)}dt+\int_{a(x+h)}^{a(x)}e^{h(x+h,t)}dt$$ $$=\int_{a(x)}^{b(x)}e^{h(x+h,t)}-e^{h(x,t)}dt+\int_{b(x)}^{b(x+h)}e^{h(x+h,t)}dt-\int_{a(x)}^{a(x+h)}e^{h(x+h,t)}dt$$ So that $$\frac{F(x+h)-F(x)}h=\int_{a(x)}^{b(x)}\frac{e^{h(x+h,t)}-e^{h(x,t)}}hdt+\frac1h\int_{b(x)}^{b(x+h)}e^{h(x+h,t)}dt-\frac1h\int_{a(x)}^{a(x+h)}e^{h(x+h,t)}dt$$ and the limit as $h\to 0$ (when the different limits exist!) should be (perhaps... ;-)) : $$F'(x)=\int_{a(x)}^{b(x)}\frac{\partial h(x,t)}{\partial x}.e^{h(x,t)}dt+b'(x)e^{h(x,b(x))}-a'(x)e^{h(x,a(x))}$$

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