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!
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!
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 $f(-x)$.
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.