Derivative of definite integrals with interval variables inside the integral i.e.: $ \frac{d}{dx} \int_a^x{(x-t)^n f^{(n+1)}(t)}dt $

Question

How do I differentiate definite integrals with the variable of differentiation as the upper or lower limit of integration / inside the integral such as the one below? $$ \frac{d}{dx} \int_a^x{(x-t)^n f^{(n+1)}(t)}dt $$ I know that integrating regular integrals i.e.: $ \frac{d}{dx} \int_a^x{f(t)}dt =f(x) $ Here, $x$ is the upper bound but $x$ is also within $ (x-t)^n $ so what happens to that? $$ $$ Edit: is it possible to have a general solution to this? Fundamental theorem of calculus says $ \frac{d}{dx} \int_a^x{f(t)}dt =f(x) $ but what about $ \frac{d}{dx} \int_a^x{f(t, x)}dt $ ? $$ $$ Edit2: obviously $$ \frac{d}{dx} \int_0^x {(x + t)^2}dt = \frac{d}{dx} \left(\frac{(x+t)^3}{3}\big|^x_0\right) = \frac{d}{dx}\frac{7x^3}{3} = 7x^2$$ but $$ \frac{d}{dx} \int_0^x {(x + t)^2}dt \not= (x+t)^2\big|^x_0 = 3x^2$$

Answer 1

Use integration by parts $n$ times. I'll get you started: let $u=(x-t)^n$ and $dv=f^{(n+1)}(t)\,dt$. Then $du=-n(x-t)^{n-1}$ and $v=f^{(n)}(t)$. Thus, $$\frac{d}{dx}\int_0^x(x-t)^nf^{(n+1)}(t)\,dt=\frac{d}{dx}\left((x-t)^nf^{(n)}(t)\big|^x_a+n\int_0^x (x-t)^{n-1}f^{(n)}(t)\,dt\right)$$ $$=-n(x-a)^{n-1}f^{(n)}(a)+n\frac{d}{dx}\int_0^x (x-t)^{n-1}f^{(n)}(t)\,dt$$

now do this $n-1$ more times (Hint: induction is your friend.)