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I know that $\frac{d}{dx} \int_{0}^{x} f(t) dt = f(x)$. What about $\frac{d}{dx} \int_{0}^{x} f(t-x) dt$? Is that just $f(x-x)$? I think I have to use the chain rule but I'm not sure how.

Thank you!

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    @RagibZaman: Thank you for the link!2011-11-16

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The following is an informal way of figuring out what the answer is. For someone like me who has a limited number of brain cells, it beats trying to remember a formula.

Let $F(u)$ be any antiderivative (indefinite integral) of $f(u)$. Then $\int_0^x f(t-x)\,dt=\left.F(t-x)\right|_0^x=F(0)-F(-x).$ Now differentiate with respect to $x$. By the definition of antiderivative, we have F'(u)=f(u). Thus, by the Chain Rule, our derivative is $-(-f(-x))$, or more simply $f(-x)$.

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    Wow that is clever :D Thank you very much!2011-11-16
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Another approach would be the following:

$\int\limits_0^x {f\left( {t - x} \right)dt} \overbrace = ^{t - x = u}\int\limits_{ - x}^0 {f\left( u \right)du} = - \int\limits_0^{ - x} {f\left( u \right)du} $

From this it is immediate that

$\frac d{dx}\int\limits_0^x {f\left( {x - t} \right)dt} =-\frac d{dx} \int\limits_0^{-x} {f\left( u \right)du}=f(-x) $

where the $-$ signs cancelled.

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    @JackPenny It's a typo, it should be $t-x=u$ (a change of variables).2012-05-04