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I know how to compute the derivative of $f(x-s)$ and by the fundamental theorem of calculus the derivative of $\int_0^x f(s)\,ds$ is $f(x)$. But I can't figure out how to do it when they're mashed together as in $\int_0^x f(x-s)\,ds$. The presence of the $s$ inside $f(x-s)$ prevents me from factoring out $f$ and using the product rule.

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You use the Leibniz formula $$ \frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)\;dt = f(x,b(x))b'(x) - f(x,a(x))a'(x) + \int_{a(x)}^{b(x)}\frac{\partial}{\partial x}f(x,t)\;dt$$ and get $$ \frac{d}{dx}\int_0^xf(x-t)\;dt = f(0)+\int_0^xf'(x-t)\; dt.$$

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Another way is by a change of variable, which is also essentially how one prove the Leibniz formula mentioned by @spaceisdarkgreen . Take $u = t/x$, then we see that \begin{align} \int^x_0 f(x-t)\ dt = \int^1_0 xf(x(1- u))\ du. \end{align} Taking the derivative with respect $x$ gives us \begin{align} \frac{d}{dx}\int^x_0 f(x-t)\ dt =&\ \int^1_0 f(x(1-u))\ du+ \int^1_0 xf'(x(1-u))(1-u)\ du. \end{align} Using integration by part formula, we get \begin{align} \int^1_0 xf'(x(1-u))(1-u)\ du =&\ - f(x(1-u))(1-u)\big|^1_{u=0} - \int^1_0 f(x(1-u))\ du\\ =&\ f(x) -\int^1_0 f(x(1-u))\ du \end{align} which means \begin{align} \frac{d}{dx}\int^x_0 f(x-t)\ dt = f(x). \end{align}

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  • $\begingroup$ (+1) for noticing a massive simplification I missed and not gloating about it. $\endgroup$ – spaceisdarkgreen Nov 29 '17 at 5:02

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