# About differentiation under the product integral sign

It is known that $$\dfrac{d}{dx}\int_{a(x)}^{b(x)}f(x,t)~dt=\dfrac{db(x)}{dx}f(x,b(x))-\dfrac{da(x)}{dx}f(x,a(x))+\int_{a(x)}^{b(x)}\dfrac{\partial}{\partial x}f(x,t)~dt.$$

How about $$\dfrac{d}{dx}\prod_{a(x)}^{b(x)}f(x,t)^{dt}?$$

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Because we are in a commutative setting, the product integral satisfies

$$\prod_{a(x)}^{b(x)} f(x,t)^{dt}=\exp\left(\int_{a(x)}^{b(x)}\log f(x,t)\,dt\right).$$

Therefore with the chain rule and the rule you state, we derive

$$\frac{d}{dx}\log \left(\prod_{a(x)}^{b(x)} f(x,t)^{dt}\right) = \frac{d}{dx}\int_{a(x)}^{b(x)}\log f(x,t)dt$$ $$= b'(x)\log f(x,b(x))-a'(x)\log f(x,a(x))+\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}\log f(x,t)dt$$

hence

$$\frac{d}{dx}\prod_{a(x)}^{b(x)} f(x,t)^{dt}=\left(\prod_{a(x)}^{b(x)} f(x,t)^{dt}\right)\left(\log \frac{f(x,b(x))^{b'(x)}}{f(x,a(x))^{a'(x)}}+\int_{a(x)}^{b(x)}\frac{\frac{\partial}{\partial x}f(x,t)}{f(x,t)}dt\right).$$

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