Derivative of a convolution

Question

I need to find the derivative of the following equation, which I do think is a convolution:

Could anybody give me a hint on how to find the derivative of V(x)?

Many thanks in advance!

Answer 1

Generally, let $\exists F(x,t): F_t(x,t)=\dfrac{\partial F(x,t)}{\partial t}, F_x(x,t)=\dfrac{\partial F(x,t)}{\partial x}, F_{x,t}(x,t)=\dfrac{\partial^2 F(x,t)}{\partial x\;\partial t}$

Specifically: $F_t(x,t)=f(t)\cdot g(x-t)$

$$\displaystyle \begin{align}\because \frac{\mathrm{d}}{\mathrm{d} x}\int_a^b F_{t}(x,t) \;\mathrm{d}t & = \frac{\mathrm{d}}{\mathrm{d} x}(F(x,b)-F(x,a)) \\ & = F_x(x,b)-F_x(x,a) \\ & = \int_a^b F_{x,t}(x,t)\;\mathrm{d} t \\ & =\int_a^b \frac{\partial}{\partial x} F_t(x,t)\;\mathrm{d} t \\ \therefore \frac{\mathrm{d}}{\mathrm{d} x} \int_{-\infty}^{\infty} f(t)\cdot g(x-t)\;\mathrm{d} t & = \int_{-\infty}^{\infty} f(t)\cdot \frac{\partial\; g(x-t)}{\partial x} \;\mathrm{d} t \end{align}$$

Answer 2

Many thanks for your replies.

Two questions:

1. Can I still apply Graham's derivation although dg(x-t)/dx in my case is NOT continous?

2. According to the other comments above, the following should also be allowed (in addition to Graham's derivation):

I can't really see that 2.) makes sense, since f(t) in a convolusion by definition never is a function of x and its derivative to x would always give 0. Could somebody maybe comment on this?!